Question 15 Marks
A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission.
|
Month
|
Sale of units
|
Total commission drawn (in Rs.)
|
|
|
A
|
B
|
C
|
|
|
Jan
|
90
|
100
|
20
|
800
|
|
Feb
|
130
|
50
|
40
|
900
|
|
March
|
60
|
100
|
30
|
850
|
Find out the rates of commission on items A, B and C by using determinant method.
AnswerLet the rates of commissions on iteams A, B and C be x, y and z respectively.
Then we can express the given modal as system of linear equations
90x + 100y + 20z = 800
130x + 50y + 40z = 900
60x + 100y + 30z = 850
We will solve this using the Cramer's rule
Here,
$\text{D}=\begin{vmatrix}90&100&20\\130&50&40\\60&100&30\end{vmatrix}=\begin{vmatrix}-170&0&-60\\130&50&40\\-200&0&-50\end{vmatrix}$
$=50(8500-12000)=-175000$
$\text{D}_1=\begin{vmatrix}800&100&20\\900&50&40\\60&100&30\end{vmatrix}=\begin{vmatrix}-170&0&-60\\130&50&40\\-950&0&-50\end{vmatrix}$
$=50(50000-57000)=-350000$
$\text{D}_2=\begin{vmatrix}90&800&20\\130&900&40\\60&850&30\end{vmatrix}=\begin{vmatrix}90&800&20\\-50&-700&0\\-75&-350&0\end{vmatrix}$
$=20(17500-52500)=-700000$
$\text{D}_3=\begin{vmatrix}90&100&800\\130&50&900\\60&100&850\end{vmatrix}=\begin{vmatrix}-170&0&-1000\\130&50&900\\-200&0&-950\end{vmatrix}$
$=50(161500-200000)=-1925000$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-350000}{-175000}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-700000}{-175000}=4$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-1925000}{-175000}=11$
$\therefore$ The rates of commission of iteam A, B and C are 2%, 4% and 11% respectively.
View full question & answer→Question 25 Marks
Solve the following systems of linear equations by cramer's rule:
$x + y + z + 1 = 0,$
$ax + by + cz + d = 0,$
$a^2x + b^2y + x^2z + d^2 = 0$
AnswerThese equations can be written as
$x + y + z + 1 = 0$
$ax + by + cz + d = 0$
$a^2x + b^2y + x^2z + d^2 = 0$
$\text{D}=\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2 \end{vmatrix}$
$=\begin{vmatrix}1&0&0\\\text{a}&\text{a}-\text{b}&\text{b}-\text{c}\\\text{a}^2&\text{a}^2-\text{b}^2&\text{b}^2-\text{c}^2 \end{vmatrix} $ [Applying $C_2 \rightarrow C_1 - C_2, C_3 \rightarrow C_2 - C_3]$
Taking (b - a) and (c - a) common from $C_1$ and $C_2$, respectively, we get
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}1&0&0\\\text{a}&1&1\\\text{a}^2&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})\ ....(\text{i})$
$\text{D}_1=\begin{vmatrix}-1&1&1\\-\text{d}&\text{b}&\text{c}\\-\text{d}^2&\text{b}^2&\text{c}^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\\text{d}&\text{b}&\text{c}\\\text{d}^2&\text{b}^2&\text{c}^2\end{vmatrix}$
$\text{D}_1=-(\text{d}-\text{b})(\text{b}-\text{c})(\text{c}-\text{d})$ [Replacing a by d in eq. (i)]
$\text{D}_2=\begin{vmatrix}1&-1&1\\\text{a}&-\text{d}&\text{c}\\\text{a}^2&-\text{d}^2&\text{c}^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\\text{a}&\text{d}&\text{c}\\\text{a}^2&\text{d}^2&\text{c}^2\end{vmatrix}$
$\text{D}_2=-(\text{a}-\text{d})(\text{d}-\text{c})(\text{c}-\text{a})$
$\text{D}_3=\begin{vmatrix}1&1&-1\\\text{a}&\text{b}&-\text{d}\\\text{a}^2&\text{b}^2&-\text{d}^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\\text{a}&\text{d}&\text{d}\\\text{a}^2&\text{b}^2&\text{d}^2\end{vmatrix}$
$\text{D}_3=-(\text{a}-\text{b})(\text{b}-\text{d})(\text{d}-\text{a})$
Thus,
$\text{x}=\frac{\text{D}_1}{\text{D}}=-\frac{(\text{d}-\text{b})(\text{b}-\text{c})(\text{c}-\text{d})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=-\frac{(\text{a}-\text{d})(\text{d}-\text{c})(\text{c}-\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{(\text{a}-\text{b})(\text{b}-\text{d})(\text{d}-\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
View full question & answer→Question 35 Marks
Solve the following systems of linear equations by cramer's rule:
$6x + y - 3z = 5,$
$x + 3y - 2z = 5,$
$2x + y + 4z = 8$
AnswerLet $\text{D}=\begin{vmatrix}6&1&-3\\1&3&-2\\2&1&4\end{vmatrix}$
Expanding along $R_1$
$=6(14)-1(8)-3(-5)$
$=84-8+15=91$
Also, $\text{D}_1=\begin{vmatrix}5&1&-3\\5&-3&-2\\8&1&4\end{vmatrix}$
Expanding along $R_1$
$=5(14)-1(36)-3(-19)$
$=70-36+57=91$
Again $\text{D}_2=\begin{vmatrix}6&5&-3\\1&5&-2\\2&8&4\end{vmatrix}$
Expanding along $R_1$
$=6(36)-5(8)-3(-2)$
$=216-40+6=182$
Also $\text{D}_3=\begin{vmatrix}6&1&5\\1&3&5\\2&1&8\end{vmatrix}$
Expanding along $R_1$
$=6(19)-1(-2)+5(-5)$
$=114+2-25=91$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{91}{91}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{182}{91}=2$
Also $\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{91}{91}=1$
Hence, $x = 1, y = 2, z = 1$
View full question & answer→Question 45 Marks
Solve the following systems of linear equations by cramer's rule:
5x - 7y + z = 11,
6x - 8y - z = 15,
3x + 2y - 6z = 7
Answer$\text{D}=\begin{vmatrix}5&-7&1\\6&-8&-1\\3&2&-6 \end{vmatrix}$
$=5(48+2)+7(-36+3)+1(12+24)$
$=5(50)+7(-33)+1(36)=55$
$\text{D}_1=\begin{vmatrix}11&-7&1\\15&-8&-1\\7&2&-6 \end{vmatrix}$
$=11(48+2)+7(-90+7)+1(30+36)$
$=11(50)+7(-83)+1(86)=55$
$\text{D}_2=\begin{vmatrix}5&11&1\\6&15&-1\\3&7&-6 \end{vmatrix}$
$=5(-90+7)-11(-36+3)+1(42-45)$
$=5(-83)-11(-33)+1(-3)=-55$
$\text{D}_3=\begin{vmatrix}5&-7&11\\6&-8&15\\3&2&7 \end{vmatrix}$
$=5(-56-30)+7(42-45)+11(12+24)$
$=5(-86)+7(-3)+11(36)=-55$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{55}{55}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-55}{55}=-1$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-55}{55}=-1$
$\therefore\text{x}=1,\text{ y}=-1$ and $\text{z}=-1$
View full question & answer→Question 55 Marks
If $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2,$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin\text{ xy})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{\text{d}}{\text{dx}}(\text{x}^2)-\frac{\text{d}}{\text{dx}}(\text{y}^2)$
$\Rightarrow \cos(\text{xy})\frac{\text{d}}{\text{dx}}(\text{xy})+\Bigg\{\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}\Bigg\}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \cos(\text{xy})\Big\{\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big\}+\Bigg\{\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}(1)}{\text{x}^2}\Bigg\}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \cos(\text{xy})\Big\{\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big\}+\frac{1}{\text{x}^2}\Big(\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\Big)=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \text{x}\cos(\text{xy})\frac{\text{dy}}{\text{dx}}+\text{y}\cos(\text{xy})+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}^2}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big\{\text{x}\cos(\text{xy})+\frac{1}{\text{x}}+2\text{y}\Big\}=\frac{\text{y}}{\text{x}^2}-\text{y}\cos(\text{xy})+2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}^2\cos(\text{xy})+1+2\text{xy}}{\text{x}}\Big\}=\frac{1}{\text{x}^2}\big(\text{y}-\text{x}^2\text{y}\cos(\text{xy})+2\text{x}^2\big)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^3+\text{y}-\text{x}^2\text{y}\cos(\text{xy})}{\text{x}\big(\text{x}^2\cos(\text{xy})+1+2\text{xy}\big)}$
View full question & answer→Question 65 Marks
Solve the following systems of linear equations by cramer's rule:
$x + y = 5,$
$y + z = 3,$
$x + z = 4$
AnswerLet $\text{D}=\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}$
Expanding along $R_1$
$=1(1)-1(-1)+0(-1)$
$=1+1+0=2$
Also $\text{D}_1=\begin{vmatrix}5&1&0\\3&1&1\\4&0&1\end{vmatrix}$
Expanding along $R_1$
$=5(1)-1(-1)+0(-4)$
$=5+1+0=6$
Again $\text{D}_2=\begin{vmatrix}1&5&0\\0&3&1\\1&4&1\end{vmatrix}$
Expanding along $R_1$
$=1(-1)-5(-1)+0(-3)$
$=-1+5+0=4$
Also $\text{D}_3=\begin{vmatrix}1&1&5\\0&1&3\\1&0&4\end{vmatrix}$
$=1(4)-1(-3)+5(-1)$
$=4+3-5=2$
Now $\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{6}{2}=3$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{4}{2}=2$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{2}{2}=1$
Hence, $x = 3, y = 2, z = 1$
View full question & answer→Question 75 Marks
Solve the following systems of linear equations by cramer's rule:
x + y = 1,
x + z = -6,
x - y - 2z = 3
AnswerThese equations can be written as
x + y + 0z = 1
x + 0y + z = -6
x - y - 2z = 3
$\text{D}=\begin{vmatrix}1&1&0\\1&0&1\\1&-1&-2 \end{vmatrix}$
$=1(0+1)-1(-2-1)+0(-1-0)=4$
$\text{D}_1=\begin{vmatrix}1&1&0\\-6&0&1\\3&-1&-2 \end{vmatrix}$
$=1(0+1)-1(12-3)+0(6-0)=-8$
$\text{D}_2=\begin{vmatrix}1&1&0\\1&-6&1\\1&3&-2 \end{vmatrix}$
$=1(12-3)-1(-2-1)+0(3+6)=12$
$\text{D}_3=\begin{vmatrix}1&1&0\\1&0&-6\\1&-1&3 \end{vmatrix}$
$=1(0-6)-1(3+6)+1(-1-0)=-16$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-8}{4}=-2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{12}{4}=3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-16}{4}=-4$
$\therefore\text{x}=-2,\text{y}=3$ and $\text{z}=-4$
View full question & answer→Question 85 Marks
Solve the following systems of linear equations by cramer's rule:
$x + y + z + w = 2,$
$x - 2y + 2z + 2w = -6,$
$2x + y - 2z + 2w = -5,$
$3x - y + 3z - 3w = -3$
AnswerHere, $\text{D}=\begin{vmatrix}1&1&1&1\\1&-2&2&2\\2&1&-2&2\\3&-1&3&-3\end{vmatrix}$
$\therefore\text{D}=\begin{vmatrix}1&0&0&0\\1&-3&1&1\\2&-1&-4&0\\3&-4&0&-6\end{vmatrix}=1\begin{vmatrix}-3&1&1\\-1&-4&0\\-4&0&-6\end{vmatrix}$
$[C_2 \rightarrow C_2 - C_1, C_3 \rightarrow C_3 - C_1, C_4 \rightarrow C_4 - C_1]$
$\therefore\text{D}=\begin{vmatrix}0&0&1\\-1&-4&0\\-22&6&-6\end{vmatrix}$ $[C_1 \rightarrow C_1 + 3C_3, C_2 \rightarrow C_2 - C_3]$
$=1(-6-88)=-94$
$\text{D}_1=\begin{vmatrix}2&1&1&1\\-6&-2&2&2\\-5&1&-2&2\\-3&-1&3&-3\end{vmatrix}=188$
$\text{D}_2=\begin{vmatrix}1&2&1&1\\1&-6&2&2\\2&-5&-2&2\\3&-3&3&-3\end{vmatrix}=-282$
$\text{D}_3=\begin{vmatrix}1&1&2&1\\1&-2&-6&2\\2&-1&-5&-2\\3&-1&-3&-3\end{vmatrix}=-141$
$\text{D}_4=\begin{vmatrix}1&1&1&2\\1&-2&2&-6\\2&1&-2&-5\\3&-1&3&-3\end{vmatrix}=47$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{188}{-94}=-2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-282}{-94}=3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-141}{94}=\frac{3}{2}$
$\text{w}=\frac{\text{D}_4}{\text{D}}=\frac{47}{-94}=-\frac{1}{2}$
Hence, $\text{x}=-2,\text{y}=3,\text{z}=\frac{3}{2},\text{w}=-\frac{1}{2}$
View full question & answer→Question 95 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
x - y + z = 3,
2x + y - z = 2,
-x - 2y + 2z = 1
AnswerWe have,
$\text{D}=\begin{vmatrix}1&-1&1\\2&1&-1\\-1&-2&2\end{vmatrix}=\begin{vmatrix}0&-1&0\\3&1&0\\-3&-2&0\end{vmatrix}=0$
$\text{D}_1=\begin{vmatrix}3&-1&1\\2&1&-1\\1&-2&2\end{vmatrix}=\begin{vmatrix}3&-1&0\\2&1&0\\1&-2&0\end{vmatrix}=0$
$\text{D}_2=\begin{vmatrix}1&3&1\\2&2&-1\\-1&1&2\end{vmatrix}=\begin{vmatrix}1&0&0\\2&-4&-3\\-1&4&3\end{vmatrix}=1(-12+12)=0$
$\text{D}_3=\begin{vmatrix}1&-1&3\\2&1&2\\-1&-2&1\end{vmatrix}=\begin{vmatrix}1&0&0\\2&3&-4\\-1&-3&4\end{vmatrix}=1(12-12)=0$
$\therefore\text{D}=\text{D}_1=\text{D}_2=\text{D}_3=0$
So, either the system is consistent with infinitely many solutions or it is inconsistent.
Consider the first two equations, written as
x - y = 3 - z
2x + y = 2 + z
to solve these equation, written as
Here,
$\text{D}=\begin{vmatrix}1&-1\\2&1\end{vmatrix}=1+2=3$
$\text{D}_1=\begin{vmatrix}3-\text{z}&-1\\2+\text{z}&1\end{vmatrix}=(3-\text{z})+(2+\text{z})=5$
$\text{D}_2=\begin{vmatrix}1&3-\text{z}\\1&2+\text{z}\end{vmatrix}=(2+\text{z})-(6-2\text{z})=-4+3\text{z}$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{5}{3}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-4+3\text{z}}{3}$
Let z = k, then the equations have the solution.
$\text{x}=\frac{5}{3},\text{ y}=\frac{-4+3\text{k}}{3},\text{ z}=\text{ k}$
View full question & answer→Question 105 Marks
Solve the following systems of linear equations by cramer's rule:
2x - 3y - 4z = 29,
-2x + 5y - z = -15,
3x - y + 5z = -11
AnswerGiven, 2x - 3y - 4z = 29
-2x + 5y - z = -15
3x - y + 5z = -11
$\text{D}=\begin{vmatrix}2&-3&-4\\-2&5&-1\\3&-1&5 \end{vmatrix}$
$=2(25-1)+3(-10+3)-4(2-15)$
$=2(24)+3(-7)-4(-13)=79$
$\text{D}_1=\begin{vmatrix}29&-3&-4\\-15&5&-1\\-11&-1&5 \end{vmatrix}$
$=29(25-1)+3(-72-11)-4(15+55)$
$=29(24)+3(-86)-4(70)=158$
$\text{D}_2=\begin{vmatrix}2&29&-4\\-2&-15&-1\\3&-11&5 \end{vmatrix}$
$=2(-75-11)-29(-10+3)-4(22+45)$
$=2(-86)-29(-7)-4(67)=-237$
$\text{D}_3=\begin{vmatrix}2&-3&29\\-2&5&-15\\3&-1&-11 \end{vmatrix}$
$=2(-55-15)+3(22+45)+29(2-15)$
$=2(-70)+3(67)+29(-13)=-316$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{158}{79}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-237}{79}=-3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-316}{79}=-4$
$\therefore\text{x}=2,\text{y}=-3$ and $\text{z}=-4$
View full question & answer→Question 115 Marks
Solve the following systems of linear equations by cramer's rule:
x - 4y - z = 11,
2x - 5y + 2z = 39,
-3x + 2y + z = 1
AnswerGiven, x - 4y - z = 11
2x - 5y + 2z = 39
-3x + 2y + z = 1
$\text{D}=\begin{vmatrix}1&-4&-1\\2&-5&2\\-3&2&1\end{vmatrix}$
$=1(-5-4)-(-4)(2+6)+(-1)(4-15)$
$=1(-9)-(-4)(8)+(-1)(-11)=34$
$\text{D}_1=\begin{vmatrix}11&-4&-1\\39&-5&2\\1&2&1\end{vmatrix}$
$=11(-5-4)-(-4)(39-2)+(-1)(78+5)$
$=11(-9)-(-4)(37)+(-1)(83)=-34$
$\text{D}_2=\begin{vmatrix}1&11&-1\\2&39&2\\-3&1&1\end{vmatrix}$
$=1(39-2)-11(2+6)+(-1)(2+117)$
$=1(37)-11(8)+(-1)(119)=-170$
$\text{D}_3=\begin{vmatrix}1&-4&11\\2&-5&39\\-3&2&1\end{vmatrix}$
$=1(-5-78)-(-4)(2+117)+11(4-15)$
$=1(-83)-(-4)(119)+11(-11)=272$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-34}{34}=-1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-170}{34}=-5$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{272}{34}=8$
$\therefore\text{x}=-1,\text{ y}=-5$ and $\text{z}=8$
View full question & answer→Question 125 Marks
Solve the following systems of linear equations by cramer's rule:
$3x + y + z = 2,$
$2x - 4y + 3z = -1,$
$4x + y - 3z = -11$
AnswerLet $\text{D}=\begin{vmatrix}3&1&1\\2&-4&3\\4&1&-3\end{vmatrix}$
Expanding along $R_1$
$=3(9)+(-1)(-18)+1(18)$
$=27+18+18=63$
Again $\text{D}_2=\begin{vmatrix}3&2&1\\2&-1&3\\4&-11&-3\end{vmatrix}$
Expanding along $R_1$
$=3(3+33)-2(-18)+1(-22+4)$
$=108+36-18=126$
Also $\text{D}_3=\begin{vmatrix}3&1&2\\2&-4&-1\\4&1&-11\end{vmatrix}$
Expanding along $R_1$
$=3(45)-1(-18)+2(18)$
$=135+18+36=189$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-63}{63}=-1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{126}{63}=2$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{189}{63}=3$
View full question & answer→Question 135 Marks
An automobile company uses three types of steel $S_1, S_2$ and $S_3$ for producing three types of cars $C_1, C_2$ and $C_3$. Steel requirements (in tons) for each type of cars are given below:
|
Steel
|
Cars
|
|
|
$C_1$
|
$C_2$
|
$C_3$
|
|
$S_1$
|
2
|
3
|
4
|
|
$S_2$
|
1
|
1
|
2
|
|
$S_3$
|
3
|
2
|
1
|
Using Cramer's rule, find the number of cars of each type which can be produced using $29, 13$ and $16$ tons of steel of three types respectively. AnswerExpressing the given information as a system of linear equations we get,
$2x +3y + 4z = 29$
$x + y + 2z = 13$
$3x + 2y + z = 16$
Where x, y, z is the number of cars $C_1, C_2$ and $C_3$ produced.
We use Cramer's rule to solve this system.
Here,
$\text{D}=\begin{vmatrix}2&3&4\\1&1&2\\3&2&1\end{vmatrix}=\begin{vmatrix}-10&-5&0\\-5&-3&0\\3&2&1\end{vmatrix}=1(30-25)=5$
$\text{D}_1=\begin{vmatrix}29&3&4\\13&1&2\\16&2&1\end{vmatrix}=\begin{vmatrix}-35&-5&0\\-19&-3&0\\16&2&1\end{vmatrix}=1(105-95)=10$
$\text{D}_2=\begin{vmatrix}0&29&4\\1&13&2\\3&16&1\end{vmatrix}=\begin{vmatrix}-10&-35&0\\-5&-19&0\\3&16&1\end{vmatrix}=(190-175)=15$
$\text{D}_3=\text{D}=\begin{vmatrix}2&3&29\\1&1&13\\3&2&16\end{vmatrix}=\begin{vmatrix}-2&0&0\\1&1&13\\3&2&16\end{vmatrix}=-2(16-26)=20$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{10}{5}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{15}{5}=3$
and $\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{20}{5}=4$
Hence, the number of cars produced of the type $C_1, C_2$ and $C_3$ are $2, 3$ and $4$ respectively.
View full question & answer→Question 145 Marks
Solve the following systems of linear equations by cramer's rule:
2x - 3z + w = 1,
x - y + 2w = 1,
-3y + z + w = 1,
x + y + z = 1
Answer$\text{D}=\begin{vmatrix}2&0&-3&1\\1&-1&0&2\\0&-3&1&1\\1&1&1&0\end{vmatrix}$
$=2\begin{vmatrix}-1&0&2\\-3&1&1\\1&1&0\end{vmatrix}-0-3\begin{vmatrix}1&-1&2\\0&-3&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&-1&0\\0&-3&1\\1&1&1\end{vmatrix}$
= 2[-1(0 - 1) - 0(0 - 1) + 2(-3 - 1)] - 3[1(0 - 1) + 1(0 - 1) + 2(0 + 3)] - 1[1(-3 - 1) + 1(0 - 1) + 0(0 + 3)]
= -21
$\text{D}_1=\begin{vmatrix}1&0&-3&1\\1&-1&0&2\\1&-3&1&1\\1&1&1&0\end{vmatrix}$
$=\begin{vmatrix}-1&0&2\\-3&1&1\\1&1&0\end{vmatrix}-0-3\begin{vmatrix}1&-1&2\\0&-3&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&-1&0\\0&-3&1\\1&1&1\end{vmatrix}$
= 2[-1(0 - 1) - 0(0 - 1) + 2(-3 - 1)] - 3[1(0 - 1) + 1(0 - 1) + 2(0 + 3)] - 1[1(-3 - 1) + 1(0 - 1) + 0(0 + 3)]
= -21
$\text{D}_2=\begin{vmatrix}2&1&3&1\\1&1&0&2\\0&1&1&1\\1&1&1&0\end{vmatrix}$
$=2\begin{vmatrix}1&0&2\\1&1&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&0&2\\0&1&1\\1&1&0\end{vmatrix}+(-3)\begin{vmatrix}1&1&2\\0&1&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&1&0\\0&1&1\\1&1&1\end{vmatrix}$
= 2[1(0 - 1) + 2(1 - 1)] - 1[1(0 - 1) + 2(0 - 1)] - 3[1(0 - 1) - 1(0 - 1) + 2(0 - 1)] - 1[1(1 - 1) - 1(0 - 1)]
= 6
$\text{D}_3=2\begin{vmatrix}-1&1&2\\-3&1&1\\1&1&0\end{vmatrix}-0+1\begin{vmatrix}1&-1&2\\0&-3&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&-1&1\\0&-3&1\\1&1&1\end{vmatrix}$
= 2[-1(0 - 1) - 1(0 - 1) + 2(-3 - 1)] + 1[1(0 - 1) + 1(0 - 1) + 2(0 + 3)] - 1[1(-3 - 1) + 1(0 - 1) + 1(0 + 3)]
= -6
$\text{D}_4=\begin{vmatrix}2&0&-3&1\\1&-1&0&1\\0&-3&1&1\\1&1&1&1\end{vmatrix}$
$=2\begin{vmatrix}-1&0&1\\-3&1&1\\1&1&1\end{vmatrix}-0-3\begin{vmatrix}1&-1&1\\0&-3&1\\1&1&1\end{vmatrix}-1\begin{vmatrix}1&-1&0\\0&-3&1\\1&1&1\end{vmatrix}$
= 2[-1(1 - 1) + 1(-3 - 1)] - 3[1(-3 - 1) + 1(0 - 1) + 1(0 + 3)] - 1[1(-3 - 1) + 1(0 - 1)]
= 3
So, by Cramer's rule, we obtain
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{21}{21}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{6}{-21}=-\frac{2}{7}$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{3}{-21}=-\frac{1}{7}$
$\text{w}=\frac{\text{D}_4}{\text{D}}=\frac{3}{-21}=-\frac{1}{7}$
Hence, $\text{x}=1,\text{ y}=-\frac{2}{7},\text{ z}=\frac{2}{7},\text{ w}=-\frac{1}{7}$
View full question & answer→Question 155 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
$2x + y - 2z = 0,$
$x - 2y + z = -2,$
$5x - 5y + z = -2$
Answer$=\begin{vmatrix}2&1&-2\\1&-2&1\\5&5&1\end{vmatrix}=\begin{vmatrix}12&9&-12\\-4&-3&1\\0&0&1\end{vmatrix}=1(-36+36)=0$
$\text{D}_1=\begin{vmatrix}4&1&-2\\-2&-2&1\\-2&-5&1\end{vmatrix}=\begin{vmatrix}0&1&-2\\0&-2&1\\0&-5&1\end{vmatrix}=0$
$\text{D}_2=\begin{vmatrix}2&4&-2\\1&-2&1\\5&-2&1\end{vmatrix}=\begin{vmatrix}2&0&-2\\1&0&1\\5&0&1\end{vmatrix}=0$
$\text{D}_3=\begin{vmatrix}2&1&4\\1&-2&-2\\5&-5&-2\end{vmatrix}=\begin{vmatrix}4&-3&0\\1&-2&-2\\4&-3&0\end{vmatrix}=2(-12+12)=0$
So, $\text{D}=\text{D}_1=\text{D}_2=\text{D}_3=0$
So, the given system is either inconsistent or has infinite solutions.
Consider the $2^{nd}$ and $3^{rd}$ equations, written as
$x - 2y = -2 - z$
$5x - 5y = -2 - z$
Then,
$\text{D}=\begin{vmatrix}1&-2\\5&-5\end{vmatrix}=-5-(-10)=5$
$\text{D}_2=\begin{vmatrix}-2-\text{z}&-2\\-2-\text{z}&-5\end{vmatrix}$
$=(2+\text{z})+(5)-2(2+\text{z})=3(2+\text{z})=6+3\text{z}$
$\text{D}_2=\begin{vmatrix}1&-(2+\text{z})\\5&-(2+\text{z})\end{vmatrix}$
$=-(2+\text{z})+5(2+\text{z})=4(2+\text{z})=8+4\text{z}$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{6+3\text{z}}{5}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{8+4\text{z}}{5}$
Let z = k, then
$\text{x}=\frac{6+3\text{k}}{5},\text{ y}=\frac{8+4\text{k}}{5},\text{ z}=\text{k}$ are the infinite solution of the given system of equations.
View full question & answer→Question 165 Marks
Solve the following systems of linear equations by cramer's rule:
2y - 3z = 0,
x + 3y = -4,
3x + 4y = 3
AnswerThese equations can be written as
0x + 2y - 3z = 0
x + 3y + 0z = -4
3x + 4y + 0z= 3
$\text{D}=\begin{vmatrix}0&2&-3\\1&3&0\\3&4&0 \end{vmatrix}$
$=0(0-0)-2(0-0)-3(4-9)$
$=15$
$\text{D}_1=\begin{vmatrix}0&2&-3\\-4&3&0\\3&4&0 \end{vmatrix}$
$=0(0-0)-0(0-0)-3(3+12)$
$=-45$
$\text{D}_3=\begin{vmatrix}0&2&0\\1&3&-4\\3&4&3 \end{vmatrix}$
$=0(9+16)-2(3+12)-0(4-9)$
$=-30$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{75}{15}=5$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-45}{15}=-3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-30}{12}=-2$
$\therefore\text{x}=5,\text{y}=-3$ and $\text{z}=-2$
View full question & answer→Question 175 Marks
Solve the following determinant equations:
$\begin{vmatrix}\text{x}+1&3&5\\2&\text{x}+2&5\\2&3&\text{x}+4\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}\text{x}+1&3&5\\2&\text{x}+2&5\\2&3&\text{x}+4\end{vmatrix}$
$=\begin{vmatrix}\text{x}+9&3&5\\\text{x}+9&\text{x}+2&5\\\text{x}+9&3&\text{x}+4\end{vmatrix}$ [Applying $C_1 = C_1 + C_2 + C_3]$
$=(\text{x}+9)\begin{vmatrix}1&3&5\\1&\text{x}+2&5\\1&3&\text{x}+4\end{vmatrix}$
$=(\text{x}+9)\begin{vmatrix}1&3&5\\0&\text{x}-1&0\\1&3&\text{x}+4\end{vmatrix}=0$ [Applying $R_2 \rightarrow R_2 - R_1]$
$=(\text{x}+9)\begin{vmatrix}1&3&5\\0&\text{x}-1&0\\1&0&\text{x}-1\end{vmatrix}$ [Applying $R_3 \rightarrow R_3 - R_1]$
$=(\text{x}+9)(\text{x}-1)^2=0$
$\text{x}=-9,1,1$
View full question & answer→Question 185 Marks
Prove that:
$\begin{vmatrix}\text{a}+\text{b}+2\text{c}&\text{a}&\text{b}\\\text{c}&\text{b}+\text{c}+2\text{a}&\text{b}\\\text{c}&\text{a}&\text{c}+\text{a}+2\text{b} \end{vmatrix}=2(\text{a}+\text{b}+\text{c})^3$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}+2\text{c}&\text{a}&\text{b}\\\text{c}&\text{b}+\text{c}+2\text{a}&\text{b}\\\text{c}&\text{a}&\text{c}+\text{a}+2\text{b} \end{vmatrix}$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{a}&\text{b}\\1&\text{b}+\text{c}+2\text{a}&\text{b}\\1&\text{a}&\text{c}+\text{a}+2\text{b} \end{vmatrix}$ [Taking out 2(a + b + c) common from $C_1]$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{a}&\text{b}\\1&\text{b}+\text{c}+\text{a}&0\\0&-\text{b}-\text{c}-\text{a}&\text{c}+\text{a}+\text{b} \end{vmatrix}$ [Applying $R_2 \rightarrow R_2 - R_1$ and $R_2 \rightarrow R_2 - R_3]$
$=2(\text{a}+\text{b}+\text{c})(\text{a}+\text{b}+\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{a}&\text{b}\\0&1&0\\0&-1&0\end{vmatrix}$ [Taking out (a + b + c) common from $R_2$ and $R_3]$
$=2(\text{a}+\text{b}+\text{c})^3\{1(1-0)\}$ [Expanding along $C_1]$
$=2(\text{a}+\text{b}+\text{c})^3$
$=\text{R.H.S}$
View full question & answer→Question 195 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin^223^{\circ}&\sin^267^{\circ}&\cos180^{\circ}\\-\sin^267^{\circ}&-\sin^223^{\circ}&\cos^2180^{\circ}\\\cos180^{\circ}&\sin^223^{\circ}&\sin^267^{\circ}\end{vmatrix}$
Answer$\begin{vmatrix}\sin^223^{\circ}&\sin^267^{\circ}&\cos180^{\circ}\\-\sin^267^{\circ}&-\sin^223^{\circ}&\cos^2180^{\circ}\\\cos180^{\circ}&\sin^223^{\circ}&\sin^267^{\circ}\end{vmatrix}$
$=\begin{vmatrix}\sin^223^{\circ}&\sin^2(90-23)^{\circ}&-1\\-\sin^2(90-23)^{\circ}&-\sin^223^{\circ}&1\\1&\sin^223^{\circ}&\sin^2(90-23)^{\circ}\end{vmatrix}$
$=\begin{vmatrix}\sin^223^{\circ}&\cos^223^{\circ}&-1\\-\cos^223^{\circ}&-\sin^223^{\circ}&1\\-1&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=\begin{vmatrix}\sin^223^{\circ}+\cos^223^{\circ}&\cos^223^{\circ}&-1\\-\cos^223^{\circ}-\sin^223^{\circ}&-\sin^223^{\circ}&1\\-1+\sin^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$ [Applying $C_1 \rightarrow C_1 + C_2]$
$=\begin{vmatrix}1&1&-1\\-1&-\sin^223^{\circ}&1\\-\cos^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=(-1)\begin{vmatrix}1&1&-1\\-1&-\sin^223^{\circ}&1\\-\cos^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=0$
View full question & answer→Question 205 Marks
Prove that:
$\begin{vmatrix}\text{a}^2&\text{a}^2-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&\text{b}^2-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&\text{c}^2-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2&\text{a}^2-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&\text{b}^2-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&\text{c}^2-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=\begin{vmatrix}\text{a}^2&-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$ [Applying $C_2 \rightarrow C_2 - C1]$
$=(-1)\begin{vmatrix}\text{a}^2&(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=-\begin{vmatrix}\text{a}^2&\text{b}^2+\text{c}^2&\text{bc}\\\text{b}^2&\text{c}^2+\text{a}^2&\text{ca}\\\text{c}^2&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix}$ [Applying $C_2 \rightarrow C_2 - 2C_1]$
$=-\begin{vmatrix}\text{a}^2+\text{b}^2+\text{c}^2&\text{b}^2+\text{c}^2&\text{bc}\\\text{b}^2+\text{c}^2+\text{a}^2&\text{c}^2+\text{a}^2&\text{ca}\\\text{c}^2+\text{a}^2+\text{b}&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix}$ [Applying $C_1 \rightarrow C_1 + C_2]$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\1&\text{c}^2+\text{a}^2&\text{ca}\\1&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix}$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\0&\text{a}^2-\text{b}^2&\text{c}(\text{a}-\text{b})\\0&\text{a}^2-\text{c}^2&\text{b}(\text{a}-\text{c})\end{vmatrix}$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{a}-\text{c})\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\0&\text{a}+\text{b}&\text{c}\\0&\text{a}+\text{c}&\text{b}\end{vmatrix} $
[Taking (a - b) common from $R_2$ and (a - c) common from $R_3]$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})\times\left\{1\times\begin{vmatrix}\text{a}+\text{b}&\text{c}\\\text{a}+\text{c}&\text{b}\end{vmatrix}\right\}$
$[\because(\text{c}-\text{a})=-(\text{a}-\text{c})]$ [Expanding along $C_1$]
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})(\text{ab}+\text{b}^2-\text{ac}-\text{c}^2)$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})\{\text{a}(\text{b}-\text{c})+(\text{b}+\text{c})(\text{b}-\text{c})\}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$
Hence prove.
View full question & answer→Question 215 Marks
Evaluate:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{c}&\text{a}&\text{b}\\\text{b}&\text{c}&\text{a}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{c}&\text{a}&\text{b}\\\text{b}&\text{c}&\text{a}\end{vmatrix}$
Applying $C_1 \rightarrow C_1 + C_2 + C_3$ we get,
$\triangle=\begin{vmatrix}\text{a}+\text{b}+\text{c}&\text{b}&\text{c}\\\text{a}+\text{b}+\text{c}&\text{a}&\text{b}\\\text{a}+\text{b}+\text{c}&\text{c}&\text{a}\end{vmatrix}$
Taking (a + b + c) common, we have
$\triangle=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\1&\text{a}&\text{b}\\1&\text{c}&\text{a}\end{vmatrix}$
Applying $R_2 \rightarrow R_2 - R_1, R_3 - R_1$, we get
$\triangle=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\0&\text{a}-\text{b}&\text{b}-\text{c}\\0&\text{c}-\text{b}&\text{a}-\text{c}\end{vmatrix}$
$\Rightarrow\triangle=(\text{a}+\text{b}+\text{c})[(\text{a}-\text{b})(\text{a}-\text{c})-(\text{b}-\text{c})(\text{c}-\text{b})]$
$\Rightarrow\triangle=(\text{a}+\text{b}+\text{c})\big[\text{a}^2-\text{ac}-\text{ab}+\text{bc}+\text{b}^2+\text{c}^2-2\text{ab}\big]$
$\Rightarrow\triangle=(\text{a}+\text{b}+\text{c})\big[\text{a}^2+\text{b}^2+\text{c}^2-\text{ac}-\text{ab}-\text{bc}\big]$
View full question & answer→Question 225 Marks
Prove that: $\begin{vmatrix}(a+1)(a+2)&(a+2)&1\\(a+2)(a+3)&(a+3)&1\\(a+3)(a+4)&(a+4) &1\end{vmatrix}=-2$
Answer$\begin{vmatrix}(a+1)(a+2)&a+2&1\\(a+2)(a+3)&a+3&1\\(a+3)(a+4)&a+4 &1\end{vmatrix}=-2$
$\text{L.H.S}=\begin{vmatrix}(a+1)(a+2)&a+2&1\\(a+2)(a+3)&a+3&1\\(a+3)(a+4)&a+4 &1\end{vmatrix}$ Apply R3 → R3 - R2
$=\begin{vmatrix}(a+1)(a+2)&a+2&1\\(a+2)(a+3)&a+3&1\\(a+3)2&1&0\end{vmatrix}$ Apply R2 → R2 - R1
$=\begin{vmatrix}(a+1)(a+2)&a+2&1\\(a+2)2&1&0\\(a+3)2&1&0\end{vmatrix}$
$=[(2a+4)(1)-(1)(2a+6)]$
$=-2$
$=\text{R.H.S}$
View full question & answer→Question 235 Marks
Evaluate:
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
When a = b, the first two rows become identical. Hence, a - b is a factor. Similarly, when b = c and c = a, the second and third and third and first rows become indetical. Hence, b - c and c - a are also factors. The degree of product of the diagonal elements is 3. Hence, there are no other factors.
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
$=\lambda(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$ [Where $\lambda$ is a constant]
$\begin{vmatrix}1&0&2\\1&1&0\\1&2&0\end{vmatrix}=2\lambda$ $[$Putting a = 0, b = 1 and c = 2 to find $\lambda]$
$\Rightarrow2=2\lambda$
$\Rightarrow\lambda=1$
Hence,
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$
View full question & answer→Question 245 Marks
Prove that:
$\begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\2&3+2\text{p}&4+3\text{p}+2\text{p}\\3&6+3\text{p}&10+6\text{p}+3\text{q}\end{vmatrix}=1$
Answer$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+\begin{vmatrix}1&1&\text{p}\\2&3&3\text{p}\\3&6&6\text{p}\end{vmatrix}+(\text{pq})\begin{vmatrix}1&1&1\\2&2&2\\3&3&3\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+(\text{p})\begin{vmatrix}1&1&\text{p}\\2&3&3\\3&6&6\end{vmatrix}+0$
$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+0$
$[\because$ Value of determinant with two identical columns is zero$]$
$=\begin{vmatrix}1&0&0\\2&1&2\\3&3&7\end{vmatrix}$ [Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=\left\{1\times\begin{vmatrix}1&2\\3&7\end{vmatrix}\right\}$ [Expanding along $R_1]$
$=7-6$
$=1$
$=\text{R.H.S}$
View full question & answer→Question 255 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
Answer$\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
$=\begin{vmatrix}\sin\alpha\sin\delta&\cos\alpha\cos\delta&\cos(\alpha+\delta)\\\sin\beta\sin\delta&\cos\beta\cos\delta&\cos(\beta+\delta)\\\sin\gamma\sin\delta&\cos\gamma\cos\delta&\cos(\gamma+\delta)\end{vmatrix}$$[\text{Applying} \text{ C}_1\rightarrow\sin\delta\text{ C}_1\text{ and}\text{ C}_2\rightarrow\cos\delta\text{ C}_2]$
$=\begin{vmatrix}\sin\alpha\sin\delta&\cos(\alpha+\delta)&\cos(\alpha+\delta)\\\sin\beta\sin\delta&\cos(\beta+\delta)&\cos(\beta+\delta)\\\sin\gamma\sin\delta&\cos(\gamma+\delta)&\cos(\gamma+\delta)\end{vmatrix}$ [Applying $C_2 \rightarrow C_2 - C_1]$
$=0$
View full question & answer→Question 265 Marks
Prove the following identities:
$\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=(5\text{x}+\lambda)(\lambda-\text{x})^2$
Answer$\text{L.H.S}=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}-\text{x}-\lambda&\text{x}+\lambda-2\text{x}&0\\2\text{x}-\text{x}-\lambda&0&\text{x}+\lambda-2\text{x}\end{vmatrix}$ [Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1]$
$=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\-(\lambda-\text{x})&\lambda-\text{x}&0\\-(\lambda-\text{x})&0&\lambda-\text{x}\end{vmatrix}$
$=(\lambda-\text{x})^2\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\-1&1&0\\-1&0&1\end{vmatrix}$ [Taking $(\lambda-\text{x})$ common from $R_2$ and $(\lambda-\text{x})$ common from $R_3]$
$=(\lambda-\text{x})^2[-1(-2\text{x})+1(\text{x}+\lambda+2\text{x})]$ [Expanding along last row]
$=(\lambda-\text{x})^2(\lambda+5\text{x})$
$=\text{R.H.S}$
$\because\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=(\lambda-\text{x})^2(\lambda+5\text{x})$
View full question & answer→Question 275 Marks
Prove the following identities:
$\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}=4\text{xyz}$
Answer$\text{L.H.S}=\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
Applying $R_1 \rightarrow R_1 - R_2$
$=\begin{vmatrix}\text{y}&-\text{x}&\text{y}-\text{x}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
Applying $R_1 \rightarrow R_1 - R_3$
$=\begin{vmatrix}0&-2\text{x}&-2\text{x}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
$=2\text{x}[\text{z}(\text{x}+\text{y})-\text{xy}]-2\text{x}[\text{zx}-\text{y}(\text{z}+\text{x})]$
$=2\text{x}[\text{zx}+\text{zy}-\text{xy}-\text{zx}+\text{yz}+\text{yx}]$
$=4\text{xyz}$
View full question & answer→Question 285 Marks
Prove that:
$\begin{vmatrix}\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}=2\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}$
Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants, we get
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}+(-1)\begin{vmatrix}\text{a}&\text{c}&\text{b}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{b}&\text{a} \end{vmatrix}$
$[$Applying $\text{C}_1\leftrightarrow\text{C}_3$ in second determinant to get negative value of the deteminant$]$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}+(-1)(-1)\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}$ $[$Applying $\text{C}_2\leftrightarrow\text{C}_3]$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}=\text{R.H.S}$
View full question & answer→Question 295 Marks
Solve the following systems of linear equations by cramer's rule:
3x + ay = 4,
2x + ay = 2, $\text{a}\neq0$
AnswerGiven, 3x + ay = 4
2x + ay = 2
Using Cramer's rule, we get
$\text{D}=\begin{vmatrix}3&\text{a}\\2&\text{a}\end{vmatrix}=3\text{a}-2\text{a}=\text{a}$
$\text{D}_1=\begin{vmatrix}4&\text{a}\\2&\text{a}\end{vmatrix}=4\text{a}-2\text{a}=2\text{a}$
$\text{D}_2=\begin{vmatrix}3&4\\2&2\end{vmatrix}=6-8=-2$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{2\text{a}}{\text{a}}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-2}{\text{a}}=-\frac{2}{\text{a}}$
$\therefore\text{x}=2$ and $\text{y}=-\frac{2}{\text{a}}$
View full question & answer→Question 305 Marks
Using properties of determinants prove that:
$\begin{vmatrix}\text{x}+4&2\text{x}&2\text{x}\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$
$=(5\text{x}+4)(4-\text{x})^2$
Answer$\text{L.H.S}=\begin{vmatrix}\text{x}+4&2\text{x}&2\text{x}\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$
$=\begin{vmatrix}5\text{x}+4&5\text{x}+4&5\text{x}+4\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 + R_2 + R_3]$
$=5\text{x}+4\begin{vmatrix}1&1&1\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$ [Take out 5x + 4 common from $R_1]$
$=5\text{x}+4\begin{vmatrix}1&0&0\\2\text{x}&4-\text{x}&0\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$ [Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=5\text{x}+4(4-\text{x})^2$
View full question & answer→Question 315 Marks
If the points (x, -2), (5, 2), (8, 8) are collinear, find x using determinants.
AnswerThe points $(k, -2), (5, 2), (8, 8)$ are collinear.
$\begin{vmatrix}\text{x}&-2&1\\5&2&1\\8&8&1\end{vmatrix}=0$
$\triangle=\begin{vmatrix}\text{x}&-2&1\\5&2&1\\8&8&1\end{vmatrix}$
$=\begin{vmatrix}\text{x}&-2&1\\5-\text{x}&4&0\\8&8&1\end{vmatrix}$ [Applying $R_2 \rightarrow R_2 - R_1]$
$=\begin{vmatrix}\text{x}&-2&1\\5-\text{x}&4&0\\8-\text{x}&10&0\end{vmatrix}$ [Applying $R_3 \rightarrow R_3 - R_1]$
$=\begin{vmatrix}5-\text{x}&4\\8-\text{x}&10\end{vmatrix}$
$=50-10\text{x}-32+4\text{x}$
$=18-6\text{x}=0$
$\Rightarrow\text{x}=3$
View full question & answer→Question 325 Marks
Evaluate:
$\begin{vmatrix}\text{x}+\lambda&\text{x}&\text{x}\\\text{x}&\text{x}+\lambda&\text{x}\\\text{x}&\text{x}&\text{x}+\lambda\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\text{x}+\lambda&\text{x}&\text{x}\\\text{x}&\text{x}+\lambda&\text{x}\\\text{x}&\text{x}&\text{x}+\lambda\end{vmatrix}$
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&\lambda&\text{x}\\0&-\lambda&\text{x}+\lambda\end{vmatrix}$ [Applying $C_1 \rightarrow C_1 - C_2, C_2 \rightarrow C_2 - C_3]$
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&0&2\text{x}+\lambda\\0&-\lambda&\text{x}+\lambda\end{vmatrix}$ [Applying $R_1 \rightarrow R_2 + R_3]$
$=\lambda\begin{vmatrix}0&2\text{x}+\lambda\\-\lambda&\text{x}+\lambda\end{vmatrix}+\text{x}\begin{vmatrix}-\lambda&0\\0&-\lambda\end{vmatrix}$
$=\lambda\big[\lambda(2\text{x}+\lambda)\big]+\text{x}\lambda^2$
$=\lambda^2(2\text{x}+\lambda+\lambda^2\text{x})$
$=3\lambda^2\text{x}+\lambda^3$
$=\lambda^2(3\text{x}+\lambda)$
View full question & answer→Question 335 Marks
$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha \end{vmatrix}$
AnswerGiven,
$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha \end{vmatrix}$
$\Rightarrow\triangle=(-1)^{1+1}\cos\alpha\cos\beta(\cos\alpha\cos\beta-0)\\+(-1)^{1+2}\cos\alpha\sin\beta(-\sin\beta\cos\alpha-0)\\+(-1)^{1+3}(-\sin\alpha)(-\sin^2\beta\sin\alpha-\sin\alpha\cos^2\beta)$ [Expanding along $R_1]$
$=\cos\alpha\cos\beta(\cos\alpha\cos\beta-0)-\cos\alpha\sin\beta(-\sin\beta\cos\alpha-0)\\-\sin\alpha(\sin^2\beta\sin\alpha-\sin\alpha-\sin\alpha\cos^2\beta)$
$=\cos^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta+\sin^2\alpha\sin^2\beta+\sin^2\alpha\cos^2\beta$
$=\cos^2\alpha(\cos^2\beta+\sin^\beta)+\sin^2\alpha(\sin^2\beta+\cos^2\beta)$
$\Rightarrow\triangle=\cos^2\alpha+\sin^2\alpha$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$\Rightarrow\triangle=1$
View full question & answer→Question 345 Marks
$\begin{vmatrix}\text{a}+\text{b}+\text{c}&-\text{c}&-\text{b}\\-\text{c}&\text{a}+\text{b}+\text{c}&-\text{a}\\-\text{b}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$
$=2(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
Answer$\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}+\text{c}&-\text{c}&-\text{b}\\-\text{c}&\text{a}+\text{b}+\text{c}&-\text{a}\\-\text{b}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&-\text{c}&-\text{b}\\\text{b}&\text{a}+\text{b}+\text{c}&-\text{a}\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ [Applying $C_1 \rightarrow C_1 + C_2 + C_3]$
$=\begin{vmatrix}\text{a}+\text{b}&\text{a}+\text{b}&-(\text{a}+\text{b})\\\text{b}+\text{c}&\text{b}+\text{c}&\text{b}+\text{c}\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 + R_2$and $R_2 \rightarrow R_2 + R_3]$
$=(\text{a}+\text{b})(\text{b}+\text{c})\begin{vmatrix}1&1&-1\\1&1&1\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$
[Taking out common factor from $R_1$ and $R_2]$
$=(\text{a}+\text{b})(\text{b}+\text{c})\begin{vmatrix}0&0&-2\\1&1&1\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ [Applying $R_1 \rightarrow R_1- R_2]$
$=(\text{a}+\text{b})(\text{b}+\text{c})\{(-2)(-\text{a}-\text{c})\}$ [Expanding along $R_1]$
$=2(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
$=\text{R.H.S}$
Hence proved.
View full question & answer→Question 355 Marks
Without expanding, prove that:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\\\text{a}&\text{b}&\text{c}\end{vmatrix}=\begin{vmatrix}\text{y}&\text{b}&\text{q}\\\text{x}&\text{a}&\text{p}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$
Answer$=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\\\text{a}&\text{b}&\text{c}\end{vmatrix}$ $\text{R}_2\leftrightarrow\text{R}_3$
$=-\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$ $\text{R}_1\leftrightarrow\text{R}_2$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$
$\begin{vmatrix}\text{y}&\text{b}&\text{q}\\\text{x}&\text{a}&\text{p}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$
$=\begin{vmatrix}\text{y}&\text{x}&\text{z}\\\text{b}&\text{a}&\text{c}\\\text{q}&\text{p}&\text{r}\end{vmatrix}$ $\text{C}_1\leftrightarrow\text{C}_2$
$=-\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$ $\text{R}_1\leftrightarrow\text{R}_2$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$
View full question & answer→Question 365 Marks
Show that the following systems of linear equations is inconsistent:
$3x - y + 2z = 3,$
$2x + y + 3z = 5,$
$x - 2y - z = 1$
AnswerGiven,
$3x - y + 2z = 3,$
$2x + y + 3z = 5,$
$x - 2y - z = 1$
$\text{D}=\begin{vmatrix}3&-1&2\\2&1&3\\1&-2&-1\end{vmatrix}$
$=3(-1+6)+1(-2-3)+2(-4-1)=0$
$\text{D}_1=\begin{vmatrix}3&-1&2\\5&1&3\\1&-2&-1\end{vmatrix}$
$=3(-1+6)+1(-5-3)+2(-10-1)=-15$
$\text{D}_2=\begin{vmatrix}3&3&2\\2&5&3\\1&1&-1\end{vmatrix}$
$=3(-5-3)-3(-2-3)+2(2-5)=-15$
$\text{D}_3=\begin{vmatrix}3&-1&3\\2&1&5\\1&-2&1\end{vmatrix}$
$=3(1+10)+1(2-5)+3(-4-1)=-15$
Here, d is zero, but $D_1, D_2$ and$ D_3$ are non-zero. Thus, the system of linear equations is inconsistent.
View full question & answer→Question 375 Marks
Solve the following determinant equations:
$\begin{vmatrix}3&-2&\sin(3\theta)\\-7&8&\cos(2\theta)\\-11&14&2\end{vmatrix}=0$
Answer$\begin{vmatrix}3&-2&\sin3\theta\\-7&8&\cos2\theta\\-11&14&2\end{vmatrix}=0$
$\Rightarrow3(16-14\cos2\theta)+2(-14+11\cos2\theta)\\+\sin3\theta(-98+88)=0$
$\Rightarrow20(1-\cos2\theta)+10\sin3\theta=0$
$\Rightarrow20(2\sin^2\theta)+10(3\sin\theta-4\sin^3\theta)=0$
$\Rightarrow4\sin^2\theta+3\sin\theta-4\sin^3\theta=0$
$\Rightarrow4\sin^2\theta+3-4\sin^2\theta=0$
$\Rightarrow4\sin^2\theta-4\sin\theta-3=0$
$\Rightarrow(2\sin\theta+1)(2\sin\theta-3)=0$
$\Rightarrow\sin\theta=-\frac{1}{2}$ or $\sin\theta=\frac{3}{2}=1.5$
As $\sin\theta\in[-1,1]$
$\therefore\sin\theta=-\frac{1}{2}$
$\Rightarrow\theta=\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{6},\text{n }\in\text{ z}$
View full question & answer→Question 385 Marks
If a, b and c are all non-zero and $\begin{vmatrix}1+\text{a}&1&1\\1&1+\text{b}&1\\1&1&1+\text{c} \end{vmatrix}=0,$ then prove that $\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}+1=0.$
Answer$\begin{vmatrix}1+\text{a}&1&1\\1&1+\text{b}&1\\1&1&1+\text{c} \end{vmatrix}=0$
$C_1 \rightarrow C_1 - C_2$
$\begin{vmatrix}\text{a}&1&1\\-\text{b}&1+\text{b}&1\\1&1&1+\text{c} \end{vmatrix}=0$
$C_2 \rightarrow C_2 - C_3$
$\begin{vmatrix}\text{a}&0&1\\-\text{b}&\text{b}&1\\0&-\text{c}&1+\text{c} \end{vmatrix}=0$
Expanding along $R_1,$ we get
$\text{a}(\text{b}+\text{bc}+\text{c})+1(\text{bc})=0$
$\Rightarrow\text{ab}+\text{abc}+\text{ac}+\text{bc}=0$
Dividing by abc, we get
$\frac{1}{\text{c}}+1+\frac{1}{\text{b}}+\frac{1}{\text{a}}=0$
$\therefore\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}+1=0$
View full question & answer→Question 395 Marks
Solve the following systems of linear equations by cramer's rule:
x + 2y = 1,
3x + y = 4
Answer$\text{D}=\begin{vmatrix}1&2\\3&1\end{vmatrix}=-5$
$\text{D}_1=\begin{vmatrix}1&2\\4&1\end{vmatrix}=-7$
$\text{D}_2=\begin{vmatrix}1&1\\3&4\end{vmatrix}=1$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{7}{5}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=-\frac{1}{5}$
$\therefore\text{x}=\frac{7}{5}$ and $\text{y}=-\frac{1}{5}$
View full question & answer→Question 405 Marks
Prove that: $\begin{vmatrix}(b+c)^2&a^2&\text{bc}\\(c+a)^2&b^2&\text{ca}\\(a+b)^2&c^2&\text{ab}\end{vmatrix}$ $=(a-b)(b-c)(c-b)(a+b+c)(a^2+b^2+c^2)$
Answer$\text{L.H.S}=\begin{vmatrix}(b+c)^2&a^2&bc\\(c+a)^2&b^2&ca\\(a+b)^2&c^2&ab\end{vmatrix}$
$=\begin{vmatrix}(b+c)^2-(c+a)^2&a^2-b^2&bc-ca\\(c+a)^2-(a+b)^2&b^2-c^2&ca-ab\\(a+b)^2&c^2&ab\end{vmatrix}$ [Applying R1 → R1 - R2 and R2 → R2 - R1]
$=\begin{vmatrix}(b+c)(b+2c+a)&(a+b)(a-b)&c(b-a)\\(c-a)(b+2a+c)&(b-c)(b+c)&a(c-b)\\(a+b)^2&c^2&ab\end{vmatrix}$
$=(a-b)(b-c)\begin{vmatrix}-(b+2c+a)&a+b&-c\\-(b+2a+c)&b+c&-a\\(a+b)^2&c^2&ab\end{vmatrix}$ [Applying $x^2 - y^2 = (x + y)(x - y)$ and taking out $(a - b)$ common from R1 and $(b - c)$ from R2]
$=(a-b)(b-c)\begin{vmatrix}-2(b+c+a)&a+b&-c\\-2(b+a+c)&b+c&-a\\(a+b)^2-c^2&c^2&ab\end{vmatrix}$ [Applying C1 → C1 - C2] $=(a-b)(b-c)\begin{vmatrix}-2(b+c+a)&a+b&-c\\-2(b+a+c)&b+c&-a\\(a+b+c)(a+b-c)&c^2&ab\end{vmatrix}$ [Applying $x^2 - y^2 = (x + y)(x - y)$ in C1] $=(a-b)(b-c)(a+b+c)\begin{vmatrix}-2&a+b&-c\\-2&b+c&-a\\(a+b-c)&c^2&ab\end{vmatrix}$ [Taking out $(a + b + c)$ common from C1] $=(a-b)(b-c)(a+b+c)\begin{vmatrix}-2&a+b&-c\\0&c-a&c-a\\(a+b-c)&c^2&ab\end{vmatrix}$ [Applying R2 → R2 - R1] $=(a-b)(b-c)(a+b+c)(c-a)\begin{vmatrix}-2&a+b&-c\\0&1&1\\(a+b-c)&c^2&ab\end{vmatrix}$ [Taking out $(c - a)$ common from R2] $=(a-b)(b-c)(a+b+c)(c-a)\begin{vmatrix}-2&a+b+c&-c\\0&0&1\\(a+b-c)&c^2-ab&ab\end{vmatrix}$ [Applying C2 → C2 - C3] $=(a-b)(b-c)(a+b+c)(c-a) \left\{(-1)\begin{vmatrix}-2&a+b+c&\\(a+b-c)&c^2-ab\end{vmatrix}\right\}$ [Expanding along R2] $=-(a-b)(b-c)(a+b+c)(c-a)\{-2c^2+2ab-a^2-b^2-2ab+c^2\}$
$=-(a-b)(b-c)(a+b+c)(c-a)(-a^2-b^2-c^2)$
$=(a-b)(b-c)(c-b)(a+b+c)(a^2+b^2+c^2)$
$=\text{R.H.S}$
View full question & answer→Question 415 Marks
Without expanding, show that the values of the following determinant are zero: $\begin{vmatrix}(2^{x}+2^{-x})^2&(2^{x}-2^{-x})^2&1\\(3^{x}+3^{-x})^2&(3^{x}-3^{-x})^2&1\\(4^{x}+4^{-x})^2&(4^{x}-4^{-x})^2&1\end{vmatrix}$
Answer$\begin{vmatrix}(2^{x}+2^{-x})^2&(2^{x}-2^{-x})^2&1\\(3^{x}+3^{-x})^2&(3^{x}-3^{-x})^2&1\\(4^{x}+4^{-x})^2&(4^{x}-4^{-x})^2&1\end{vmatrix}$
$=\begin{vmatrix}(2^{x}+2^{-x}+2)&(2^{x}-2^{-x}-2)&1\\(3^{x}+3^{-x}+2)&(3^{x}-3^{-x}-2)&1\\(4^{x}+4^{-x}+2)&(4^{x}-4^{-x}-2)&1\end{vmatrix}$
$=\begin{vmatrix}4&(2^{x}+2^{-x}-2)&1\\4&(3^{x}+3^{-x}-2)&1\\4&(4^{x}+4^{-x}-2)&1\end{vmatrix}$ [Applying C1 → C1 - C2]
$=4\begin{vmatrix}1&(2^{x}+2^{-x}-2)&1\\1&(3^{x}+3^{-x}-2)&1\\1&(4^{x}+4^{-x}-2)&1\end{vmatrix}$
$=0$
View full question & answer→Question 425 Marks
Find values of k, if area of triangle is $4$ square units whose vertices are:
$(k, 0), (4, 0), (0, 2)$
AnswerIf the area of a triangle with vertices (k, 0), (4, 0) and (0, 2) is 4 square units, then
$\triangle=\frac{1}{2}\begin{vmatrix}\text{k}&0&1\\4&0&1\\0&2&1\end{vmatrix}$
$=\frac{1}{2}\left\{(2)\times\begin{vmatrix}\text{k}&1\\4&1\end{vmatrix}\right\}$ [Expanding along $C_2]$
$=(\text{k}-4)$
Since area is always +ve, we take its absolute value, which is given as 4 square units.
$\Rightarrow(\text{k}-4)=\pm4$
$\Rightarrow (k - 4) = 4$ or $(k - 4) = -4$
$\Rightarrow k - 4 = 4$ or $k - 4 = -4$
$\Rightarrow k = 8$ or $k = 0$
$\Rightarrow k = 8, 0$
View full question & answer→Question 435 Marks
Prove that:
$\begin{vmatrix}\text{a}-\text{b}-\text{c}&2\text{a}&2\text{a}\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}=(\text{a}+\text{b}+\text{c})^3$
Answer$\begin{vmatrix}\text{a}-\text{b}-\text{c}&2\text{a}&2\text{a}\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}=(\text{a}+\text{b}+\text{c})^3$
$\text{L.H.S}=\begin{vmatrix}\text{a}-\text{b}-\text{c}&2\text{a}&2\text{a}\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}$
Apply: $R_1 \rightarrow R_1 + R_2 + R_3$
$\begin{vmatrix}\text{a}+\text{b}+\text{c}&\text{a}+\text{b}+\text{c}&\text{a}+\text{b}+\text{c}\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}$
Take (a + b + c) common from $R_1$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&1&1\\2\text{b}&\text{b}-\text{c}-\text{a}&2\text{b}\\2\text{c}&2\text{c}&\text{c}-\text{a}-\text{b} \end{vmatrix}$
Apply: $C_2 \rightarrow C_2- C_1, C_3 \rightarrow C_3 - C_1$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&0&0\\2\text{b}&-\text{b}-\text{c}-\text{a}&0\\2\text{c}&2\text{c}&-\text{c}-\text{a}-\text{b} \end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&0&0\\2\text{b}&\text{b}+\text{c}+\text{a}&0\\2\text{c}&2\text{c}&\text{c}+\text{a}+\text{b} \end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})^3$
$=\text{R.H.S}$
View full question & answer→Question 445 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}&\cos\text{x}&\sin\text{y}\\-\cos\text{x}&\sin\text{x}&-\cos\text{y} \end{vmatrix}$
Answer$\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}&\cos\text{x}&\sin\text{y}\\-\cos\text{x}&\sin\text{x}&-\cos\text{y} \end{vmatrix}$
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y})\\\sin\text{x}\sin\text{y}&\cos\text{x}\sin\text{y}&\sin^2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
[Applying $R_2 \rightarrow $ siny $R_2$ and $R_3 \rightarrow $ cosy $R_3]$
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}\sin\text{y}-\cos\text{x}\cos\text{y}&\cos\text{x}\sin\text{y}+\sin\text{x}\sin\text{y}&\sin^2\text{y}-\cos^2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
[Applying $R_2 \rightarrow R_2 + R_3]$
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
$=0$
View full question & answer→Question 455 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin^2\text{A}&\cot\text{A}&1\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}&\cot\text{C}&1\end{vmatrix}$
Answer$\begin{vmatrix}\sin^2\text{A}&\cot\text{A}&1\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}&\cot\text{C}&1\end{vmatrix}$
$=\begin{vmatrix}\sin^2\text{A}-\sin^2\text{B}&\cot\text{A}-\cot\text{B}&0\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}-\sin^2\text{B}&\cot\text{C}-\cot\text{B}&0\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 - R_2$ and $R_3 \rightarrow R_3 - R_2]$
$=\begin{vmatrix}\sin(\text{A}+\text{B})\sin(\text{A}-\text{B})&\frac{\cos\text{A}\sin\text{B}-\cos\text{B}\sin\text{A}}{\sin\text{A}\sin\text{B}}&0\\\sin^2\text{B}&\cot\text{B}&1\\\sin(\text{C}+\text{B})\sin(\text{C}-\text{B})&\frac{\cos\text{C}\sin\text{B}-\cos\text{B}\sin\text{C}}{\sin\text{B}\sin\text{C}}&0\end{vmatrix}$
$=\begin{vmatrix}\sin(\pi-\text{C})\sin(\text{A}-\text{B})&\frac{-\sin(\text{A}-\text{B})}{\sin\text{A}\sin\text{B}}&0\\\sin^2\text{B}&\cot\text{B}&1\\\sin(\pi-\text{A})\sin(\text{C}-\text{B})&\frac{-\sin(\text{C}-\text{B})}{\sin\text{B}\sin\text{C}}&0\end{vmatrix}$ $[\because\text{A}+\text{B}+\text{C}=\pi]$
$=\begin{vmatrix}\sin\text{C}\sin(\text{A}-\text{B})&\frac{-\sin(\text{A}-\text{B})}{\sin\text{A}\sin\text{B}}&0\\\sin^2\text{B}&\frac{\cos\text{B}}{\sin\text{B}}&1\\\sin\text{A}\sin(\text{C}-\text{B})&\frac{-\sin(\text{C}-\text{B})}{\sin\text{A}\sin\text{B}}&0\end{vmatrix}$
$=\frac{\sin(\text{A}-\text{B})\sin(\text{C}-\text{B})}{\sin\text{B}}\begin{vmatrix}\sin\text{C}&\frac{-1}{\sin\text{A}}&0\\\sin^2\text{B}&\cos\text{B}&1\\\sin\text{A}&\frac{-1}{\sin\text{C}}&0\end{vmatrix}$
$=\frac{\sin(\text{A}-\text{B})\sin(\text{C}-\text{B})}{\sin\text{B}\sin\text{A}\sin\text{C}}\begin{vmatrix}\sin\text{C}\sin\text{A}&-1&0\\\sin^2\text{B}&\cos\text{B}&1\\\sin\text{A}\sin\text{C}&-1&0\end{vmatrix}$ [Applying $R_1 \rightarrow $ sinA $R_1$ and $R_3 \rightarrow$ sinC $R_3]$
$=\frac{\sin(\text{A}-\text{B})\sin(\text{C}-\text{B})}{\sin\text{B}\sin\text{A}\sin\text{C}}\begin{vmatrix}0&0&0\\\sin^2\text{B}&\cos\text{B}&1\\\sin\text{A}\sin\text{C}&-1&0\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 - R_3]$
$=0$
View full question & answer→Question 465 Marks
Prove that:
$\begin{vmatrix}\text{a}^2&\text{bc}&\text{ac}+\text{c}^2\\\text{a}^2+\text{ab}&\text{b}^2&\text{ac}\\\text{ab}&\text{b}^2+\text{ac}&\text{c}^2\end{vmatrix}=4\text{a}^2\text{b}^2\text{c}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2&\text{bc}&\text{ac}+\text{c}^2\\\text{a}^2+\text{ab}&\text{b}^2&\text{ac}\\\text{ab}&\text{b}^2+\text{ac}&\text{c}^2\end{vmatrix}$
$=\text{abc}\begin{vmatrix}\text{a}&\text{c}&\text{a}+\text{c}\\\text{a}+\text{b}&\text{b}&\text{a}\\\text{b}&\text{b}+\text{c}&\text{c}\end{vmatrix}$
[Taking out a, b and c common from $C_1, C_2$ and $C_3]$
$=\text{abc}\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}+\text{b}&\text{b}&-2\text{b}\\\text{b}&\text{b}+\text{c}&-2\text{b}\end{vmatrix}$
[Applying $C_3\rightarrow C_3 - C_2 - C_1]$
$=(\text{abc})(-2\text{b})\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}+\text{b}&\text{b}&1\\\text{b}&\text{b}+\text{c}&1\end{vmatrix}$
[Taking (-2b) common from $C_3]$
$=(\text{abc})(-2\text{b})\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}&-\text{c}&0\\\text{b}&\text{b}+\text{c}&1\end{vmatrix}$
[Applying $R_2 \rightarrow R_2 - R_1]$
$=(\text{abc})(-2\text{b})\times1\begin{vmatrix}\text{a}&\text{c}\\\text{a}&-\text{c}\end{vmatrix}$
[expanding along $C_3]$
$=(\text{abc})(-2\text{b})(-2\text{ac})$
$=4\text{a}^2\text{b}^2\text{c}^2$
$=\text{R.H.S}$
View full question & answer→Question 475 Marks
Prove that:
$\begin{vmatrix}\text{a}^2&2\text{ab}&\text{b}^2\\\text{b}^2&\text{a}^2&2\text{ab}\\2\text{ab}&\text{b}^2&\text{a}^2\end{vmatrix}=(\text{a}^3+\text{b}^3)^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2&2\text{ab}&\text{b}^2\\\text{b}^2&\text{a}^2&2\text{ab}\\2\text{ab}&\text{b}^2&\text{a}^2\end{vmatrix}$
$=\text{a}^2\begin{vmatrix}\text{a}^2&2\text{ab}\\\text{b}^2&\text{a}^2\end{vmatrix}-(2\text{ab})\begin{vmatrix}\text{b}^2&2\text{ab}\\2\text{ab}&\text{a}^2\end{vmatrix}+\text{b}^2\begin{vmatrix}\text{b}^2&\text{a}^2\\2\text{ab}&\text{b}^2\end{vmatrix}$ [Expanding]
$=\text{a}^2(\text{a}^4-2\text{ab}^3)-(2\text{ab})(\text{b}^2\text{a}^2-4\text{a}^2\text{b}^2)+\text{b}^2(\text{b}^4-2\text{a}^3\text{b})$
$=\text{a}^6-2\text{a}^3\text{b}^3-2\text{a}^3\text{b}^3+8\text{a}^3\text{b}^3+\text{b}^6-2\text{a}^3\text{b}^3$
$=\text{a}^6+2\text{a}^3\text{b}^3+(\text{b}^3)^2$
$=(\text{a}^3+\text{b}^3)^2$
$=\text{R.H.S}$
View full question & answer→Question 485 Marks
Solve the following determinant equations:
$\begin{vmatrix}\text{x}+\text{a}&\text{b}&\text{c}\\\text{a}&\text{x}+\text{b}&\text{c}\\\text{a}&\text{b}&\text{x}+\text{c}\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}\text{x}+\text{a}&\text{b}&\text{c}\\\text{a}&\text{x}+\text{b}&\text{c}\\\text{a}&\text{b}&\text{x}+\text{c}\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\text{a}+\text{b}+\text{c}&\text{b}&\text{c}\\\text{x}+\text{a}+\text{b}+\text{c}&\text{x}+\text{b}&\text{c}\\\text{x}+\text{a}+\text{b}+\text{c}&\text{b}&\text{x}+\text{c}\end{vmatrix}$ [Applying $C_1 \rightarrow C_1 + C_2 + C_3]$
$=(\text{x}+\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\1&\text{x}+\text{b}&\text{c}\\1&\text{b}&\text{x}+\text{c}\end{vmatrix}$
$=(\text{x}+\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\0&\text{x}&0\\1&\text{b}&\text{x}+\text{c}\end{vmatrix}$ [Applying $R_2 \rightarrow R_2 - R_1]$
$=(\text{x}+\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\0&\text{x}&0\\1&0&\text{x}\end{vmatrix}$ [Applying $R_3 \rightarrow R_3 - R_1]$
$=(\text{x}+\text{a}+\text{b}+\text{c})(\text{x}^2-0)=0$ [Given]
$\Rightarrow\text{x}^2=0$ or $\text{x}+\text{a}+\text{b}+\text{c}=0$
$\Rightarrow\text{x}=0$ or $\text{x}=-(\text{a}+\text{b}+\text{c})$
View full question & answer→Question 495 Marks
For what value of x the matrix A is singular?
$\text{A}=\begin{vmatrix}1+\text{x}&7\\3-\text{x}&8 \end{vmatrix}$
Answer$\text{A}=\begin{vmatrix}1+\text{x}&7\\3-\text{x}&8 \end{vmatrix}$
A matrix A is called singular if |A| = 0
Now expanding along the first row |A|
$=(\text{x}-1)\begin{vmatrix}\text{x}-1&1\\1&\text{x}-1 \end{vmatrix}-1\begin{vmatrix}1&1\\1&\text{x}-1 \end{vmatrix}+1\begin{vmatrix}1&1\\\text{x}-1&-1 \end{vmatrix}$
$=(\text{x}-1)\big[(\text{x}-1)^2-1\big]-1[\text{x}-1-1]+1[1-\text{x}+1]$
$=(\text{x}-1)(\text{x}^2+1-2\text{x}-1)-1(\text{x}-2)+1(2-\text{x})$
$=(\text{x}-1)(\text{x}^2-2\text{x})-\text{x}+2+2-\text{x}$
$=(\text{x}-1)\times(\text{x})\times(\text{x}-2)+(4-2\text{x})$
$=(\text{x}-1)\times(\text{x})\times(\text{x}-2)+2(2-\text{x})$
$=(\text{x}-1)\times(\text{x})\times(\text{x}-2)-2(\text{x}-2)$
$=(\text{x}-2)[\text{x}(\text{x}-1)-2]$ (Taking (x - 2) common)
Since A is a singular matrix, so |A| = 0
i.e., $(\text{x}-2)(\text{x}^2-\text{x}-2)=0$
either $(\text{x}-2)=0$ or $\text{x}^2-\text{x}-2=0$
$\text{x}=2$ or $\text{x}^2-2\text{x}+\text{x}-2=0$
$\text{x}(\text{x}-2)+1(\text{x}-2)=0$
$(\text{x}-2)(\text{x}+1)=0$
$\text{x}=2,-1$
$\text{x}=2\text{ or}-1$
View full question & answer→Question 505 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\text{a}+\text{b}&2\text{a}+\text{b}&3\text{a}+\text{b}\\2\text{a}+\text{b}&3\text{a}+\text{b}&4\text{a}+\text{b}\\4\text{a}+\text{b}&5\text{a}+\text{b}&6\text{a}+\text{b} \end{vmatrix}$
Answer$\begin{vmatrix}\text{a}+\text{b}&2\text{a}+\text{b}&3\text{a}+\text{b}\\2\text{a}+\text{b}&3\text{a}+\text{b}&4\text{a}+\text{b}\\4\text{a}+\text{b}&5\text{a}+\text{b}&6\text{a}+\text{b} \end{vmatrix}$
Apply: $C_3 \rightarrow C_3 - C_2$
$\begin{vmatrix}\text{a}+\text{b}&2\text{a}+\text{b}&\text{a}\\2\text{a}+\text{b}&3\text{a}+\text{b}&\text{a}\\4\text{a}+\text{b}&5\text{a}+\text{b}&\text{a} \end{vmatrix}$
Apply: $C_2 \rightarrow C_2 - C_1$
$\begin{vmatrix}\text{a}+\text{b}&\text{a}&\text{a}\\2\text{a}+\text{b}&\text{a}&\text{a}\\4\text{a}+\text{b}&\text{a}&\text{a} \end{vmatrix}$
$=0$
$\because\text{C}_3=\text{C}_2$
View full question & answer→Question 515 Marks
Prove that:
$\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}=16(3\text{x}+4)$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
$=\begin{vmatrix}3\text{x}+4&3\text{x}+4&3\text{x}+4\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
[Applying $R_1 \rightarrow R_2 + R_2 + R_3]$
$=(3\text{x}+4)\begin{vmatrix}1&1&1\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
[Taking out (3x + 4) common from $R_1]$
$=(3\text{x}+4)\begin{vmatrix}1&0&0\\\text{x}&4&0\\\text{x}&0&4\end{vmatrix}$
[Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=(3\text{x}+4)(4)^2$ [Expanding along $R_1]$
$=16(3\text{x}+4)$
$=\text{R.H.S}$
View full question & answer→Question 525 Marks
Solve the following systems of linear equations by cramer's rule:
3x + y = 19,
3x - y = 23
AnswerGiven, 3x + y = 19
3x - y = 23
Using cramer's Rule, we get
$\text{D}=\begin{vmatrix}3&1\\3&-1\end{vmatrix}=-3-3=-6$
$\text{D}_1=\begin{vmatrix}19&1\\23&-1\end{vmatrix}=-19-23=-42$
$\text{D}_2=\begin{vmatrix}3&19\\3&23\end{vmatrix}=(3\times23)-(3\times19)=3\times4=12$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-42}{-6}=7$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{12}{-6}=-2$
$\therefore\text{x}=7$ and $\text{y}=-2$
View full question & answer→Question 535 Marks
Evaluate the following:
$\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
$\triangle=\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
$=\begin{vmatrix}\text{x}-1&1-\text{x}&0\\1&\text{x}&1\\0&1-\text{x}&\text{x}-1\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 - R_2$ and $R_3 \rightarrow R_3 - R_2]$
$=(\text{x}-1)^2\begin{vmatrix}1&-1&0\\1&\text{x}&1\\0&-1&1\end{vmatrix}$
$=(\text{x}-1)^2\begin{vmatrix}1&-1&0\\1&\text{x}+1&1\\0&0&1\end{vmatrix}$ [Applying $C_2 \rightarrow C_2 + C_3]$
$=(\text{x}-1)^2(\text{x}+1+1)$ [Expanding along last row]
$=(\text{x}-1)^2(\text{x}+2)$
$\therefore\triangle=(\text{x}-1)^2(\text{x}+2)$
View full question & answer→Question 545 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}2&3&7\\13&17&5\\15&20&12 \end{vmatrix}$
Answer$\begin{vmatrix}2&3&7\\13&17&5\\15&20&12 \end{vmatrix}$
Use: $R_3 \rightarrow R_3 - R_2$
$=\begin{vmatrix}2&3&7\\13&17&5\\2&3&7 \end{vmatrix}$
$=0$
$\because\text{R}_3=\text{R}_1$
View full question & answer→Question 555 Marks
Solve the following determinant equations:
$\begin{vmatrix}1&\text{x}&\text{x}^3\\1&\text{b}&\text{b}^3\\1&\text{c}&\text{c}^3\end{vmatrix}=0,\text{b}\neq\text{c}$
Answer$\Rightarrow\begin{vmatrix}1&\text{x}&\text{x}^3\\1&\text{b}-\text{x}&\text{b}^3-\text{x}^3\\1&\text{c}-\text{x}&\text{c}^3-\text{x}^3\end{vmatrix}=0$
$\Rightarrow(\text{b}-\text{x})(\text{c}-\text{x})\begin{vmatrix}1&\text{x}&\text{x}^3\\0&1&\text{b}^2+\text{x}^2+\text{bx}\\0&1&\text{c}^2+\text{x}^2+\text{cx}\end{vmatrix}=0$
$\Rightarrow(\text{b}-\text{x})(\text{c}-\text{x})\begin{vmatrix}1&\text{x}&\text{x}^3\\0&1&\text{b}^2+\text{x}^2+\text{bx}\\0&1&\text{c}^2+\text{x}^2+\text{cx}-(\text{b}^2+\text{x}^2+\text{bx})\end{vmatrix}=0$
$\Rightarrow(\text{b}-\text{x})(\text{c}-\text{x})\begin{vmatrix}1&\text{x}&\text{x}^3\\0&1&\text{b}^2+\text{x}^2+\text{bx}\\0&1&\text{c}^2-\text{b}^2+\text{cx}-\text{bx}\end{vmatrix}=0$
$\Rightarrow(\text{b}-\text{x})(\text{c}-\text{x})(\text{c}-\text{b})\begin{vmatrix}1&\text{x}&\text{x}^3\\0&1&\text{b}^2+\text{x}^2+\text{bx}\\0&0&\text{b}+\text{c}+\text{x}\end{vmatrix}=0$
$\Rightarrow(\text{b}-\text{x})(\text{c}-\text{x})(\text{c}-\text{b})(\text{b}+\text{c}+\text{x})=0$
$\Rightarrow(\text{b}-\text{x})=0,(\text{c}-\text{x})=0,(\text{b}+\text{c}+\text{x})=0$
$\Rightarrow\text{x}=\text{b},\text{x}=\text{c},\text{x}=-(\text{b}+\text{c})$
View full question & answer→Question 565 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}49&1&6\\39&7&4\\26&2&3 \end{vmatrix}$
Answer$\begin{vmatrix}49&1&6\\39&7&4\\26&2&3 \end{vmatrix}$
Applying: $C_1 \rightarrow C_1 + (-8)C_3$
$=\begin{vmatrix}1&1&6\\7&7&4\\2&2&3 \end{vmatrix}=0$
$\because\text{C}_1=\text{C}_2$
View full question & answer→Question 575 Marks
Solve the following system of homogeneous linear equations:
3x + y + z = 0,
x - 4y + 3z = 0,
2x + 5y - 2z = 0
AnswerGiven,
3x + y + z = 0,
x - 4y + 3z = 0,
2x + 5y - 2z = 0
$\text{D}=\begin{vmatrix}3&1&1\\1&-4&3\\2&5&-2\end{vmatrix}=0$
The system has infinitely many solution Puting z = k in the first two equation we get
3x + y = -k
x - 4y = -3k
Solving these equations by Cramer's rule, we get
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{\begin{vmatrix}-\text{k}&1\\-3\text{k}&-4\end{vmatrix}}{\begin{vmatrix}3&1\\1&-4\end{vmatrix}}=-\frac{7\text{k}}{13}$
$\text{y}=\frac{\text{D}_2}{\text{D}} =\frac{\begin{vmatrix}3&-\text{k}\\1&-3\text{k}\end{vmatrix}}{\begin{vmatrix}3&1\\1&-4\end{vmatrix}}=\frac{8\text{k}}{13} \text{z}=\text{k}$
$\text{z}=\text{k}$
$\Rightarrow\text{x}=-\frac{7\text{k}}{13},\text{ y}=\frac{8\text{k}}{13}$ and $\text{z}=\text{x}$
Or x = -7k, y = 8k and z = 13k
Clearly, these value satisfy the thried equation.
Thus, x = -7k, y = 8k, z = 13k $[\text{k}\in\text{R}]$
View full question & answer→Question 585 Marks
If a, b, c are real numbers such that $\begin{vmatrix}\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}=0,$ then show that either $a + b + c = 0$ or $a = b= c.$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$
$=\begin{vmatrix}2(\text{a}+\text{b}+\text{c})&2(\text{a}+\text{b}+\text{c})&2(\text{a}+\text{b}+\text{c})\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 + R_2 + R_3]$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&1&1\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&0&0\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}-\text{a}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}-\text{b}\end{vmatrix}$ [Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=2(\text{a}+\text{b}+\text{c})\left\{1\begin{vmatrix}\text{b}-\text{c}&\text{b}-\text{a}\\\text{c}-\text{a}&\text{c}-\text{b}\end{vmatrix}\right\}$
$=2(\text{a}+\text{b}+\text{c})\{(\text{b}-\text{c})(\text{c}-\text{b})-(\text{b}-\text{a})(\text{c}-\text{a})\}$
$=-2(\text{a}+\text{b}+\text{c})\{\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\}$
$=-(\text{a}+\text{b}+\text{c})\{2\text{a}^2+2\text{b}^2+2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}\}$
$=-(\text{a}+\text{b}+\text{c})\{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\}$
But $\triangle=0$ [Given]
$\Rightarrow-(\text{a}+\text{b}+\text{c})\{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\}=0$
Either,
$\Rightarrow(\text{a}+\text{b}+\text{c})=0$ or $(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2=0$
$\Rightarrow(\text{a}+\text{b}+\text{c})=0$ or $\text{a}=\text{b}=\text{c}$
Hence proved.
View full question & answer→Question 595 Marks
Solve the following systems of linear equations by cramer's rule:
5x + 7y = -2
4x + 6y = -3
AnswerLet $\text{D}=\begin{vmatrix}5&7\\4&6\end{vmatrix}=-2$
$\text{D}_1=\begin{vmatrix}-2&7\\-3&6\end{vmatrix}=9$
$\text{D}_2=\begin{vmatrix}5&-2\\4&-3\end{vmatrix}=-7$
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{9}{2}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-7}{2}$
View full question & answer→Question 605 Marks
Show that $\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}=0$ where a, b, c are in A.P.
Answer$2\text{b}=\text{a}+\text{c}$
$\text{L.H.S}=\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}$ [Applying $R_2 = 2R_2]$
$=\frac{1}{2}\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\2\text{x}+4&2\text{x}+6&2\text{x}+2\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}$
$=\frac{1}{2}\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\0&0&0\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}$ $[\because2\text{b}=\text{a}+\text{c}]$
[Applying $R_2 \rightarrow R_2 - (R_1 + R_3)]$
$=0$
$=\text{R.H.S}$
View full question & answer→Question 615 Marks
If $\begin{vmatrix}\text{a}&\text{b}-\text{y}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0,$ then using properties of determinants, find the value of $\frac{\text{a}}{\text{x}}+\frac{\text{b}}{\text{y}}+\frac{\text{c}}{\text{z}},$ where $\text{x},\text{y},\text{z}\neq0.$
Answer$\Rightarrow\begin{vmatrix}\text{a}&\text{b}-\text{y}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0$
$R_1 \rightarrow R_1 - R_2$
$\Rightarrow\begin{vmatrix}\text{x}&-\text{y}&0\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0$
$R_2\rightarrow R_2 - R_3$
$\Rightarrow\begin{vmatrix}\text{x}&-\text{y}&0\\0&\text{y}&-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0$
Expanding along first row, we get
$\text{x}(\text{yc}+\text{zb}-\text{zy})+\text{y}(0-\text{za}+\text{zx})=0$
$\Rightarrow\text{xyc}+\text{xzb}-2\text{xyz}+\text{zya}=0$
$\Rightarrow\text{xyc}+\text{xzb}-2\text{xyz}+\text{zya}=0$
Dividing by xyz, we get
$\frac{\text{c}}{\text{z}}+\frac{\text{b}}{\text{y}}-2+\frac{\text{a}}{\text{x}}=0$
$\therefore\frac{\text{a}}{\text{x}}+\frac{\text{b}}{\text{y}}+\frac{\text{c}}{\text{z}}=2$
View full question & answer→Question 625 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}0&2&6\\1&5&0\\3&7&1 \end{vmatrix}$
AnswerLet $M_{ij} $and$ C_{ij}$_ are respectively the minor and co-factor of the element $a_{ij}.$
Now,
$\text{M}_{11}=\begin{vmatrix}5&0\\7&1 \end{vmatrix}=5-0=5$
$\text{M}_{21}=\begin{vmatrix}2&6\\7&1 \end{vmatrix}=2-42=-40$
$\text{M}_{31}=\begin{vmatrix}2&6\\5&0 \end{vmatrix}=0-30=-30$
$\text{C}_{11}=(-1)^{1+1}\text{M}_{11}=5$
$\text{C}_{21}=(-1)^{2+1}\text{M}_{21}=(-1)(-40)=40$
$\text{C}_{31}=(-1)^{3+1}\text{M}_{31}=(-30)=-30$
Now, expanding the determinant along the first column.
$|\text{A}|=\text{a}_{11}\text{C}_{11}+\text{a}_{21}\text{C}_{21}+\text{a}_{31}\text{C}_{31}$
$=0\times5+1\times(40)+3\times(-30)$
$=40-90$
$=-50$
View full question & answer→Question 635 Marks
Show that:
$\begin{vmatrix}\text{y}+\text{z}&\text{x}&\text{y}\\\text{z}+\text{x}&\text{z}&\text{x}\\\text{x}+\text{y}&\text{y}&\text{z}\end{vmatrix}=(\text{x}+\text{y}+\text{z})(\text{x}-\text{z})^3$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{y}+\text{z}&\text{x}&\text{y}\\\text{z}+\text{x}&\text{z}&\text{x}\\\text{x}+\text{y}&\text{y}&\text{z}\end{vmatrix}$
$=\begin{vmatrix}2(\text{x}+\text{y}+\text{z})&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\\\text{z}+\text{x}&\text{z}&\text{x}\\\text{x}+\text{y}&\text{y}&\text{z}\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 + R_2 + R_3]$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}2&1&1\\\text{z}+\text{x}&\text{z}&\text{x}\\\text{x}+\text{y}&\text{y}&\text{z}\end{vmatrix}$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}0&1&1\\0&\text{z}&\text{x}\\\text{x}-\text{z}&\text{y}&\text{z}\end{vmatrix}$ [Applying $C_1 \rightarrow C_1 - C_2 - C_3]$
$=(\text{x}+\text{y}+\text{z})\left\{(\text{x}-\text{z})\times\begin{vmatrix}1&1\\\text{z}&\text{x}\end{vmatrix}\right\}$ [Expanding along $C_1]$
$=(\text{x}+\text{y}+\text{z})(\text{x}-\text{z})^3$
View full question & answer→Question 645 Marks
Solve the following systems of linear equations by cramer's rule:
2x - y = 17,
3x + 5y = 6
AnswerGiven, 2x - y = 17
3x + 5y = 6
Using cramers Rule, we get
$\text{D}=\begin{vmatrix}2&1\\3&5\end{vmatrix}=10+3=13$
$\text{D}_1=\begin{vmatrix}17&-1\\6&5\end{vmatrix}=85+6=91$
$\text{D}_2=\begin{vmatrix}2&17\\3&6\end{vmatrix}=12-51=-39$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{91}{13}=7$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-39}{13}=-3$
$\therefore\text{x}=7$ and $\text{y}=-3$
View full question & answer→Question 655 Marks
Solve the following systems of linear equations by cramer's rule:
2x - y = 1,
7x - 2y = -7
AnswerLet $\text{D}=\begin{vmatrix}2&-1\\7&-2 \end{vmatrix}=-4+7=3$
$\text{D}_1=\begin{vmatrix}1&-1\\-7&-2 \end{vmatrix}=-9$
$\text{D}_2=\begin{vmatrix}2&1\\7&-7\end{vmatrix}=-21$
Now, $\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-9}{3}=-3$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-21}{3}=-7$
Hence x = -3, y = -7
View full question & answer→Question 665 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}2&-1&0&1\\-3&0&1&-2\\1&1&-1&1\\2&-1&5&0 \end{vmatrix}$
Answer$\text{M}_{11}=0(0-5)-1(0+1)-2(5-1)=-1-8=-9$
$\text{M}_{21}=-1(0-5)+1(5-1)=5+4=9$
$\text{M}_{31}=-1(0+10)+1(0+1)=-10+1=-9$
$\text{M}_{41}=-1(1-2)+1(0-1)=1-1=0$
$\text{C}_{11}=(-1)^{1+1}\text{M}_{11}=-9$
$\text{C}_{21}=(-1)^{2+1}\text{M}_{21}=(-1)\times9$
$\text{C}_{31}=(-1)^{3+1}\text{M}_{31}=-9$
$\text{C}_{41}=(-1)^{4+1}\text{M}_{41}=0$
$\text{D}=2\begin{vmatrix}0&1&-2\\1&-1&1\\-1&5&0 \end{vmatrix}+1\begin{vmatrix}-3&1&-2\\1&-1&1\\2&5&0\end{vmatrix}-1\begin{vmatrix}-3&0&1\\1&1&-1\\2&-1&5\end{vmatrix}$
$=-18-27+15=-30$
View full question & answer→Question 675 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab} \end{vmatrix}$
Answer$\text{M}_{11}=\begin{vmatrix}\text{b} &\text{ca}\\ \text{c}&\text{ab} \end{vmatrix}=\text{ab}^{2}-\text{c}^2\text{a}=\text{a}(\text{b}^2-\text{c}^2)$
$\text{M}_{21}=\begin{vmatrix}\text{a} &\text{bc}\\ \text{c}&\text{ab} \end{vmatrix}=\text{a}^{2}\text{b}-\text{c}^2\text{b}=\text{b}(\text{a}^2-\text{c}^2)$
$\text{M}_{31}=\begin{vmatrix}\text{a} &\text{bc}\\ \text{b}&\text{ca} \end{vmatrix}=\text{a}^{2}\text{c}-\text{b}^2\text{c}=\text{c}(\text{a}^2-\text{b}^2)$
$\text{C}_{11}=(-1)^{1+1}\text{M}_{11}=\text{a}(\text{b}^2-\text{c}^2)$
$\text{C}_{21}=(-1)^{2+1}\text{M}_{21}=-\text{b}(\text{a}^2-\text{c}^2)$
$\text{C}_{31}=(-1)^{3+1}\text{M}_{31}=\text{c}(\text{a}^2-\text{b}^2)$
$\text{D}=1\text{a}(\text{b}^2-\text{c}^2)-\text{a}(\text{ab}-\text{ca})+\text{b}(\text{c}-\text{b})$
$=\text{ab}^2-\text{ac}^2-\text{a}^2\text{b}+\text{a}^2\text{c}+\text{c}^2\text{b}-\text{b}^2\text{c}$
$=\text{a}^{2}(\text{c}-\text{b})+\text{b}^{2}(\text{a}-\text{c})+\text{c}^{2}(\text{b}-\text{a})$
View full question & answer→Question 685 Marks
Evaluate $\begin{vmatrix}2&3&-5\\7&1&-2\\-3&4&1\end{vmatrix}$ by two methods.
AnswerWe will evaluate the given determinant:
- Along the first row.
$|\text{A}|=2\begin{vmatrix}1&-2\\4&1 \end{vmatrix}-3\begin{vmatrix}7&-2\\-3&1\end{vmatrix}-3\begin{vmatrix}7&1\\-3&4\end{vmatrix}$
$=2(1+8)-7(3+20)-3(-6+5)$
$=18-7(23)-3(-1)$
$=21-161$
$=-140$
- Along the first column.
$|\text{A}|=\begin{vmatrix}1&-2\\4&1 \end{vmatrix}-7\begin{vmatrix}3&-5\\4&1\end{vmatrix}-3\begin{vmatrix}3&-5\\1&-2\end{vmatrix}$
$=2(1+8)-7(3+20)-3(-6+5)$
$=18-7(23)-3(-1)$
$=18-161+3$
$=21-161$
$=-140$
We can see, the answer is same with both the methods. View full question & answer→Question 695 Marks
$\begin{vmatrix}1&\text{a}&\text{a}^2\\\text{a}^2&1&\text{a}\\\text{a}&\text{a}^2&1\end{vmatrix}=(\text{a}^3-1)^2$
AnswerLet us consider the L.H.S of the given equation.
Let $\begin{vmatrix}1&\text{a}&\text{a}^2\\\text{a}^2&1&\text{a}\\\text{a}&\text{a}^2&1\end{vmatrix}$
Applying $C_1 \rightarrow C_1 + C_2 + C_3,$ we have,
$=\begin{vmatrix}1+\text{a}+\text{a}^2&\text{a}&\text{a}^2\\1+\text{a}+\text{a}^2&1&\text{a}\\1+\text{a}+\text{a}^2&\text{a}^2&1\end{vmatrix}$
Taking term$ (1 + a + a_2)$ common, we have,
$=(1+\text{a}+\text{a}^2)\begin{vmatrix}1&\text{a}&\text{a}^2\\1&1&\text{a}\\1&\text{a}^2&1\end{vmatrix}$
Applying $R_2 \rightarrow R_2 - R_1 $and$ R_3 \rightarrow R_3 - R_1$, we have,
$=(1+\text{a}+\text{a}^2)\begin{vmatrix}1&\text{a}&\text{a}^2\\0&1-\text{a}&\text{a}(1-\text{a})\\0&-\text{a}(1-\text{a})&(1-\text{a})(1+\text{a})\end{vmatrix}$
Taking the term (1 - a) common from $R_2 $and $R_3$, we have,
$=(1+\text{a}+\text{a}^2)(1-\text{a}^2)\begin{vmatrix}1&\text{a}&\text{a}^2\\0&1&\text{a}\\0&-\text{a}&(1+\text{a})\end{vmatrix}$
$=(1+\text{a}+\text{a}^2)(1-\text{a}^2)(1+\text{a}+\text{a}^2)$
$=(1+\text{a}+\text{a}^2)(1-\text{a}^2)$
$=\big[(1+\text{a}+\text{a}^2)(1-\text{a}^2)\big]^2$
$=\big[(\text{a}^3-1)\big]^2$
$=\text{R.H.S}$
View full question & answer→Question 705 Marks
Evaluate:
$\begin{vmatrix}\text{a}&\text{b}+\text{c}&\text{a}^2\\\text{b}&\text{c}+\text{a}&\text{b}^2\\\text{c}&\text{a}+\text{b}&\text{c}^2\end{vmatrix}$
Answer$\begin{vmatrix}\text{a}&\text{b}+\text{c}&\text{a}^2\\\text{b}&\text{c}+\text{a}&\text{b}^2\\\text{c}&\text{a}+\text{b}&\text{c}^2\end{vmatrix}$
Apply: $C_2 \rightarrow C_2 + C_1$
$=\begin{vmatrix}\text{a}&\text{b}+\text{c}+\text{a}&\text{a}^2\\\text{b}&\text{c}+\text{a}+\text{b}&\text{b}^2\\\text{c}&\text{a}+\text{b}+\text{c}&\text{c}^2\end{vmatrix}$
Take $(a + b + c)$ common from $C_2$
$=(\text{b}+\text{c}+\text{a})\begin{vmatrix}\text{a}&1&\text{a}^2\\\text{b}&1&\text{b}^2\\\text{c}&1&\text{c}^2\end{vmatrix}$
Apply: $R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - R_1$
$=(\text{b}+\text{c}+\text{a})\begin{vmatrix}\text{a}&1&\text{a}^2\\\text{b}-\text{a}&0&\text{b}^2-\text{a}^2\\\text{c}-\text{a}&0&\text{c}^2-\text{a}^2\end{vmatrix}$
$=(\text{b}+\text{c}+\text{a})(\text{b}-\text{a})(\text{c}-\text{a})\begin{vmatrix}\text{a}&1&\text{a}^2\\1&0&\text{b}+\text{a}\\1&0&\text{c}+\text{a}\end{vmatrix}$
$=(\text{b}+\text{c}+\text{a})(\text{b}-\text{a})(\text{c}-\text{a})(\text{b}-\text{c})$
View full question & answer→Question 715 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}+2\text{x}&\text{b}+2\text{y}&\text{c}+2\text{z}\\\text{x}&\text{y}&\text{z}\\\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}+2\text{x}&\text{b}+2\text{y}&\text{c}+2\text{z}\\\text{x}&\text{y}&\text{z}\\\end{vmatrix}$
$=\begin{vmatrix}\text{a}+2\text{x}&\text{b}+2\text{y}&\text{c}+2\text{z}\\\text{a}+2\text{x}&\text{b}+2\text{y}&\text{c}+2\text{z}\\\text{x}&\text{y}&\text{z}\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 + 2R_3]$
$=\begin{vmatrix}0&0&0\\\text{a}+2\text{x}&\text{b}+2\text{y}&\text{c}+2\text{z}\\\text{x}&\text{y}&\text{z}\end{vmatrix}=0$ [Applying $R_1 \rightarrow R_1 + R_2]$
View full question & answer→Question 725 Marks
Solve the following determinant equations:
$\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}=0,\text{a}\neq\text{b}$
AnswerLet $\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{x}-\text{a}&\text{x}^2-\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}$ [Applying $R_2 \rightarrow R_1 - R_2]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{x}-\text{a}&\text{x}^2-\text{a}^2\\0&\text{x}-\text{b}&\text{x}^2-\text{b}^2\end{vmatrix}$ [Applying $R_3 \rightarrow R_1 - R_3]$
$=(\text{x}-\text{a})(\text{x}-\text{b})\begin{vmatrix}1&\text{x}&\text{x}^2\\0&1&\text{x}+\text{a}\\0&1&\text{x}+\text{b}\end{vmatrix}$
$=(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}+\text{b}-\text{x}-\text{a})=0$
$\text{x}=\text{a},\text{b}$
View full question & answer→Question 735 Marks
$\begin{vmatrix}-\text{a}(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^3&2\text{c}^3\\2\text{a}^3&-\text{b}(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^3\\2\text{a}^3&2\text{b}^3&-\text{c}(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{a}^2+\text{b}^2+\text{c}^2)$
Answer$\begin{vmatrix}-\text{a}(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^3&2\text{c}^3\\2\text{a}^3&-\text{b}(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^3\\2\text{a}^3&2\text{b}^3&-\text{c}(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{a}^2+\text{b}^2+\text{c}^2)$
$\text{L.H.S}=\begin{vmatrix}-\text{a}(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^3&2\text{c}^3\\2\text{a}^3&-\text{b}(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^3\\2\text{a}^3&2\text{b}^3&-\text{c}(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
Tkae a, b and c common from $C_1, C_2$ and $C_3 $respectively.
$=\text{abc}\begin{vmatrix}-(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^2&2\text{c}^2\\2\text{a}^2&-(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^2\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
Apply: $R_1 \rightarrow R_1 - R_3, R_2 \rightarrow R_2 - R_3$
$=\text{abc}\begin{vmatrix}-(\text{b}^2+\text{c}^2-\text{a}^2)-2\text{a}^2&0&2\text{c}^2+(\text{a}^2+\text{b}^2-\text{c}^2)\\0&-(\text{c}^2+\text{a}^2-\text{b}^2)-2\text{b}^2&2\text{c}^2+(\text{a}^2+\text{b}^2-\text{c}^2)\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}\begin{vmatrix}-(\text{b}^2+\text{c}^2-\text{a}^2)-2\text{a}^2&0&(\text{a}^2+\text{b}^2-\text{c}^2)\\0&-(\text{c}^2+\text{a}^2-\text{b}^2)-2\text{b}^2&2\text{c}^2+(\text{a}^2+\text{b}^2-\text{c}^2)\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{b}^2+\text{c}^2+\text{a}^2)^2\begin{vmatrix}-1&0&1\\0&-1&1\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{b}^2+\text{c}^2+\text{a}^2)^2\begin{vmatrix}-1&0&1\\0&-1&1\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)+2\text{a}^2\end{vmatrix}$
$=\text{abc}(\text{b}^2+\text{c}^2+\text{a}^2)^2\begin{vmatrix}-1&0&1\\0&-1&1\\2\text{a}^2&2\text{b}^2&-\text{b}^2+\text{c}^2+\text{a}^2\end{vmatrix}$
$=-\text{abc}(\text{b}^2+\text{c}^2+\text{a}^2)[(-1)(-\text{b}^2+\text{c}^2+\text{a}^2)-(1)(2\text{b}^2)]$
$=\text{abc}(\text{a}^2+\text{b}^2+\text{c}^2)^3$
$=\text{R.H.S}$
View full question & answer→Question 745 Marks
Prove that:
$\begin{vmatrix}-\text{bc}&\text{b}^2+\text{bc}&\text{c}^2+\text{bc}\\\text{a}^2+\text{ac}&-\text{ac}&\text{c}^2+\text{ac}\\\text{a}^2+\text{ab}&\text{b}^2+\text{ab}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3$
Answer$\text{L.H.S}=\begin{vmatrix}-\text{bc}&\text{b}^2+\text{bc}&\text{c}^2+\text{bc}\\\text{a}^2+\text{ac}&-\text{ac}&\text{c}^2+\text{ac}\\\text{a}^2+\text{ab}&\text{b}^2+\text{ab}&-\text{ab}\end{vmatrix}$
Multiply $R_1, R_2$ and $R_3$ by a, b and c respectively.
$=\frac{1}{\text{abc}}\begin{vmatrix}-\text{abc}&\text{ab}^2+\text{abc}&\text{ac}^2+\text{abc}\\\text{a}^2\text{b}+\text{abc}&-\text{abc}&\text{bc}^2+\text{abc}\\\text{a}^2\text{c}+\text{abc}&\text{b}^2\text{c}+\text{abc}&-\text{abc}\end{vmatrix}$
Take a, b and c common from $C_1, C_2$ and $C_3$ respectively.
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}-\text{bc}&\text{ab}+\text{ac}&\text{ac}+\text{ab}\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
Apply: $R_1 \rightarrow R_1 + R_2 + R_3$
$=\begin{vmatrix}\text{ab}+\text{bc}+\text{ca}&\text{ab}+\text{bc}+\text{ca}&\text{ab}+\text{bc}+\text{ca}\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})\begin{vmatrix}1&1&1\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})\begin{vmatrix}0&1&0\\\text{a}\text{b}+\text{bc}+\text{ac}&-\text{ac}&\text{bc}+\text{ab}+\text{ac}\\0&\text{b}\text{c}+\text{ac}&-\text{ab}-\text{bc}-\text{ac}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3\begin{vmatrix}0&1&0\\0&-\text{ac}&1\\0&\text{b}\text{c}+\text{ac}&1\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3$
$=\text{R.H.S}$
View full question & answer→Question 755 Marks
$\begin{vmatrix}1+\text{a}&1&1\\1&1+\text{a}&\text{a}\\1&1&1+\text{a}\end{vmatrix}=\text{a}^3+3\text{a}^2$
Answer$\text{L.H.S}=\begin{vmatrix}1+\text{a}&1&1\\1&1+\text{a}&\text{a}\\1&1&1+\text{a}\end{vmatrix}$
$=1+\text{a}\begin{vmatrix}1+\text{a}&1\\1&1+\text{a}\end{vmatrix}-1\begin{vmatrix}1&1\\1&1+\text{a}\end{vmatrix}+1\begin{vmatrix}1&1+\text{a}\\1&1\end{vmatrix}$
$=(1+\text{a})[(1+\text{a})^2-1]-1(1+\text{a}-1)+(1-1-\text{a})$
$=(1+\text{a})[1+\text{a}^2+2\text{a}-1]-\text{a}-\text{a}$
$=1+\text{a}+\text{a}^2+\text{a}^3+2\text{a}+2\text{a}^2-2\text{a}$
$=\text{a}^3+3\text{a}^2$
$=\text{R.H.S}$
View full question & answer→Question 765 Marks
If $\triangle=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix},$ $\triangle_1=\begin{vmatrix}1&1&1\\\text{yz}&\text{zx}&\text{xy}\\\text{x}&\text{y}&\text{z}\end{vmatrix},$ then prove that $\triangle+\triangle_1=0$
Answer$\triangle+\triangle_1=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}+\begin{vmatrix}1&1&1\\\text{yz}&\text{zx}&\text{xy}\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}+\begin{vmatrix}1&\text{yz}&\text{x}\\1&\text{zx}&\text{y}\\1&\text{xy}&\text{z} \end{vmatrix}$
$[$Interchanging rows and coloumns in $\triangle_1]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}-\begin{vmatrix}1&\text{x}&\text{yz}\\1&\text{y}&\text{zx}\\1&\text{z}&\text{xy} \end{vmatrix}$
$[$Applying $\text{C}_2\leftrightarrow\text{C}_3$ in $\triangle_1]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{y}-\text{x}&\text{y}^2-\text{x}^2\\0&\text{z}-\text{x}&\text{z}^2-\text{x}^2\end{vmatrix}-\begin{vmatrix}1&\text{x}&\text{yz}\\0&\text{y}-\text{x}&\text{zx}-\text{yz}\\0&\text{z}-\text{x}&\text{xy}-\text{yz}\end{vmatrix}$
[Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1]$
$=(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}1&\text{x}&\text{x}^2\\0&1&\text{y}+\text{x}\\0&1&\text{z}+\text{x}\end{vmatrix}-(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}1&\text{x}&\text{yz}\\0&1&-\text{z}\\0&1&-\text{y}\end{vmatrix}$
[Taking (y - x) common from $R_2$ and (z - x) common from $R_3$]
$=(\text{y}-\text{x})(\text{z}-\text{x})(\text{z}+\text{x}-\text{y}-\text{x})-(\text{y}-\text{x})(\text{z}-\text{x})(-\text{y}+\text{z})$
[Expanding along first column]
$=(\text{y}-\text{x})(\text{z}-\text{x})(\text{z}-\text{y})(1-1)$
$=0$
$\therefore\ \triangle+\triangle_1=0$
View full question & answer→Question 775 Marks
Evaluate the following determinant:
$\begin{vmatrix}1&3&9&27\\3&9&27&1\\9&27&1&3\\27&1&3&9 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&3&9&27\\3&9&27&1\\9&27&1&3\\27&1&3&9 \end{vmatrix}$
$=1\begin{vmatrix}9&27&1\\27&1&3\\1&3&9 \end{vmatrix}-3\begin{vmatrix}3&27&1\\9&1&3\\27&3&6\end{vmatrix}\\+9\begin{vmatrix}3&9&1\\9&27&3\\27&1&9 \end{vmatrix}-27\begin{vmatrix}3&9&27\\9&27&1\\27&1&3 \end{vmatrix}$
$=1\{9(9-9)-27(243-3)+1(81-1)\}-3\{3(9-9)-27(81-81)+1(27-27)\}\\+9\{3(243-3)-9(81-81)+1(9-729)\}-27\{(81-1)-9(27-27)+27(9-729)\}$
$=1\{0-6480+80\}-3\{0-0+0\}+9\{720-0-720\}-27\{80-0-19440\}$
$=-6400+522720$
$=516320$
View full question & answer→Question 785 Marks
Find values of k, if area of triangle is 4 square units whose vertices are:
$(-2, 0), (0, 4), (0, k)$
Answer$4=\frac{1}{2}\begin{vmatrix}-2&0&1\\0&4&1\\0&\text{k}&1 \end{vmatrix}$
$\pm8=\begin{vmatrix}-2&0&1\\0&4&1\\0&\text{k}&1 \end{vmatrix}$
Expanding along $R_1$
$\pm8=-2(4-\text{k})-0(0-0)+1(0)$
$\pm=8=-8+2\text{k}$
Taking positive (+) sign
$\pm=8=-8+2\text{k}$
Taking positive (-) sing
$-8 = -8 + 2k$ or $k = 0$
Hence $k = 0, 8$
View full question & answer→Question 795 Marks
Prove that:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\end{vmatrix}=\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}$
Answer$\text{L.H.S}=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{a}+\text{b}+\text{c}&\text{c}+\text{a}+\text{b}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ [Applying $R_3 \rightarrow R_3 + R_2]$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\1&1&1\end{vmatrix}$ [Taking $(a + b + c)$ common]
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\1&1&1\end{vmatrix}$ [Applying $R_2 \rightarrow R_1 - R_2]$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}-\text{b}&\text{b}-\text{c}&\text{c}\\\text{b}-\text{c}&\text{c}-\text{a}&\text{a}\\0&0&1\end{vmatrix}$ $[C_1 \rightarrow C_1 - C_2$ and $C_2 \rightarrow C_2 - C_3]$
$=(\text{a}+\text{b}+\text{c})\big[-1\{(\text{a}-\text{b})(\text{c}-\text{a})-(\text{b}-\text{c})^2\}\big]$
$=(\text{a}+\text{b}+\text{c})\big[-\{\text{ac}-\text{bc}-\text{a}^2+\text{ab}-\text{b}^2-\text{c}^2+2\text{bc}\}\big]$
$=(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca})$
$=\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}=\text{ R.H.S}$
View full question & answer→Question 805 Marks
Prove that:
$\begin{vmatrix}\text{a}&\text{b}-\text{c}&\text{c}-\text{b}\\\text{a}-\text{c}&\text{b}&\text{c}-\text{a}\\\text{a}-\text{b}&\text{b}-\text{a}&\text{c}\end{vmatrix}$
$=(\text{a}+\text{b}-\text{c})(\text{b}+\text{c}-\text{a})(\text{c}+\text{a}-\text{b})$
Answer$\text{L.H.S}=\begin{vmatrix}\text{a}&\text{b}-\text{c}&\text{c}-\text{b}\\\text{a}-\text{c}&\text{b}&\text{c}-\text{a}\\\text{a}-\text{b}&\text{b}-\text{a}&\text{c}\end{vmatrix}$
$R_1 \rightarrow R_1- R_2 - R_3$
$=\begin{vmatrix}-\text{a}+\text{c}+\text{b}&-\text{b}-\text{c}+\text{a}&-\text{c}-\text{b}+\text{a}\\\text{a}-\text{c}&\text{b}&\text{c}-\text{a}\\\text{a}-\text{b}&\text{b}-\text{a}&\text{c}\end{vmatrix}$
$=(\text{b}+\text{c}-\text{a})\begin{vmatrix}1&-1&1\\\text{a}-\text{c}&\text{b}&\text{c}-\text{a}\\\text{a}-\text{b}&\text{b}-\text{a}&\text{c}\end{vmatrix}$
$=(\text{b}+\text{c}-\text{a})\begin{vmatrix}1&0&0\\\text{a}-\text{c}&\text{b}+\text{a}-\text{c}&0\\\text{a}-\text{b}&0&\text{c}+\text{a}-\text{b}\end{vmatrix}$
$=(\text{a}+\text{b}-\text{c})(\text{b}+\text{c}-\text{a})(\text{c}+\text{a}-\text{b})$
$=\text{R.H.S}$
View full question & answer→Question 815 Marks
Prove that:
$\begin{vmatrix} 1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}=\begin{vmatrix} 1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$
Answer$\begin{vmatrix} 1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
Apply $R_1 \rightarrow R_1a, R_2 \rightarrow R_2b, R_3 \rightarrow R_3c$
$=\frac{1}{\text{abc}}\begin{vmatrix} \text{a}&\text{a}^2&\text{abc}\\\text{b}&\text{b}^2&\text{cab}\\\text{c}&\text{c}^2&\text{abc}\end{vmatrix}$
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix} \text{a}&\text{a}^2&1\\\text{b}&\text{b}^2&1\\\text{c}&\text{c}^2&1\end{vmatrix}$
$=-\begin{vmatrix} \text{a}&1&\text{a}^2\\\text{b}&1&\text{b}^2\\\text{c}&1&\text{c}^2\end{vmatrix}$
$=\begin{vmatrix} 1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$
View full question & answer→Question 825 Marks
Evaluate the following determinant:
$\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
$=6(-2-10)-(-3)(4+20)+(10-10)$
$=-72+72+0$
$=0$
View full question & answer→Question 835 Marks
Solve the following determinant equations:
$\begin{vmatrix}15-2\text{x}&11-3\text{x}&7-\text{x}\\11&17&14\\10&16&13\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}15-2\text{x}&11-3\text{x}&7-\text{x}\\11&17&14\\10&16&13\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}15-2\text{x}-14+2\text{x}&11-3\text{x}&7-\text{x}\\11-28&17&14\\10-26&16&13\end{vmatrix}=0$ [Applying $C_1 \rightarrow C_1 - 2C_3]$
$\Rightarrow\begin{vmatrix}1&11-3\text{x}&7-\text{x}\\-17&17&14\\-16&16&13\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}12-3\text{x}&4-2\text{x}&7-\text{x}\\0&3&14\\0&3&13\end{vmatrix}=0$ [Applying $C_1 \rightarrow C_1 + C_2$ and $C_2 \rightarrow C_2 - C_3]$
$\Rightarrow(12-3\text{x})((3)\times13-(3\times14))=0$
$\Rightarrow(12-3\text{x})(-3)=0$
$\Rightarrow12-3\text{x}=0$
$\Rightarrow3\text{x}=12$
$\Rightarrow\text{x}=4$
View full question & answer→Question 845 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}1&43&6\\7&35&4\\3&17&2\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&43&6\\7&35&4\\3&17&2\end{vmatrix}$
$=\begin{vmatrix}1&1&6\\7&7&4\\3&3&2\end{vmatrix}=0$ [Appliying $C_2 \rightarrow C_2 - 7C_3]$
View full question & answer→Question 855 Marks
Show that x = 2 is a root of the equation $\begin{vmatrix}\text{x}&-6&-1\\2&-3\text{x}&\text{x}-3\\-3&2\text{x}&\text{x}+2\end{vmatrix}=0$ and solve it completely.
AnswerLet us show that x = 2 is a root of the given equation:
Putting x = 2 in the L.H.S, we get
$\begin{vmatrix}2&-6&-1\\2&-6&-1\\-3&4&4 \end{vmatrix}=0$
$\because\text{R}_1=\text{R}_2$
Hence, x = 2 is a root of the given equation.
Now, we see if there are any other roots. For this we need to solve the equation:
$\begin{vmatrix}\text{x}&-6&-1\\2&-3\text{x}&\text{x}-3\\-3&2\text{x}&\text{x}+2\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-1&-6&-1\\\text{x}-1&-3\text{x}&\text{x}-3\\\text{x}-1&2\text{x}&\text{x}+2\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)\begin{vmatrix}1&-6&-1\\1&-3\text{x}&\text{x}-3\\1&2\text{x}&\text{x}+2\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)\begin{vmatrix}1&-6&-1\\0&-3\text{x}+6&\text{x}-3+1\\0&2\text{x}+6&\text{x}+2+1\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)\begin{vmatrix}1&-6&-1\\0&-3(\text{x}-2)&\text{x}-2\\0&2(\text{x}+3)&\text{x}+3\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)(\text{x}-2)(\text{x}+3)\begin{vmatrix}1&-6&-1\\0&-3&1\\0&2&1 \end{vmatrix}=0\\$
$\Rightarrow(\text{x}-1)(\text{x}-2)(\text{x}+3)=0$
$\Rightarrow(\text{x}-1)=0,(\text{x}-2)=0,(\text{x}+3)=0$
$\Rightarrow\text{x}=1,\text{x}=2,\text{x}=-3$
View full question & answer→Question 865 Marks
Solve the following systems of linear equations by cramer's rule:
9x + 5y = 10,
3x - 2y = 8
AnswerGiven, 9x + 5y = 10
3y - 2x = 8
Rearranging the second equation, the two equations can be written as
9x + 5y = 10
-2x + 3y = 8
Now,
$\text{D}=\begin{vmatrix}9&5\\-2&3\end{vmatrix}=27+10=37$
$\text{D}_1=\begin{vmatrix}10&5\\8&3\end{vmatrix}=30-40=-10$
$\text{D}_2=\begin{vmatrix}9&10\\-2&8\end{vmatrix}=72+20=-92$
Using Cramer's rule we get
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-10}{37}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{92}{37}$
$\therefore\text{x}=\frac{-10}{37}$ and $\text{y}=\frac{92}{37}$
View full question & answer→Question 875 Marks
Solve the following determinant equations:
$\begin{vmatrix}\text{x}+\text{a}&\text{x}&\text{x}\\\text{x}&\text{x}+\text{a}&\text{x}\\\text{x}&\text{x}&\text{x}+\text{a}\end{vmatrix}=0,\text{a}\neq0$
AnswerLet $\begin{vmatrix}\text{x}+\text{a}&\text{x}&\text{x}\\\text{x}&\text{x}+\text{a}&\text{x}\\\text{x}&\text{x}&\text{x}+\text{a}\end{vmatrix}$
$=\begin{vmatrix}3\text{x}+\text{a}&\text{x}&\text{x}\\3\text{x}+\text{a}&\text{x}+\text{a}&\text{x}\\3\text{x}+\text{a}&\text{x}&\text{x}+\text{a}\end{vmatrix}$ [Applying $C_1 \rightarrow C_1 + C_2 + C_3]$
$=(3\text{x}+\text{a})\begin{vmatrix}1&\text{x}&\text{x}\\1&\text{x}+\text{a}&\text{x}\\1&\text{x}&\text{x}+\text{a}\end{vmatrix}$
$=(3\text{x}+\text{a})\begin{vmatrix}1&\text{x}&\text{x}\\0&\text{a}&0\\1&\text{x}&\text{x}+\text{a}\end{vmatrix}$ [Applying $R_2 \rightarrow R_2 - R_1]$
$=(3\text{x}+\text{a})\begin{vmatrix}1&\text{x}&\text{x}\\0&\text{a}&0\\1&0&\text{a}\end{vmatrix}$ [Applying $R_3 \rightarrow R_3 - R_1]$
$=(3\text{x}+\text{a})=(\text{a}^2-0)=0$
$\text{x}=\frac{-\text{a}}{3}$
View full question & answer→Question 885 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}1&\text{a}&\text{a}^2-\text{bc}\\1&\text{b}&\text{b}^2-\text{ac}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&\text{a}&\text{a}^2-\text{bc}\\1&\text{b}&\text{b}^2-\text{ac}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
$=\begin{vmatrix}0&\text{a}-\text{b}&\text{a}^2-\text{bc}-\text{b}^2+\text{ac}\\0&\text{b}-\text{c}&\text{b}^2-\text{ac}-\text{c}^2+\text{ab}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$ [applying $R_1 \rightarrow R_1 - R_2, R_2 \rightarrow R_2 - R_3]$
$=\begin{vmatrix}0&\text{a}-\text{b}&(\text{a}-\text{b})(\text{a}+\text{b})+\text{c}(\text{a}-\text{b})\\0&\text{b}-\text{c}&(\text{b}-\text{c})(\text{b}+\text{c})+\text{a}(\text{b}-\text{c})\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}0&1&(\text{a}+\text{b}+\text{c})\\0&1&(\text{a}+\text{b}+\text{c})\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
$\Rightarrow\triangle=0$
View full question & answer→Question 895 Marks
If $\begin{vmatrix}\text{p}&\text{q}&\text{c}\\\text{a}&\text{q}&\text{c}\\\text{a}&\text{b}&\text{r} \end{vmatrix}=0,$ find the value of $\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}},$ $\text{p}\neq\text{a},\text{q}\neq\text{b},\text{r}\neq\text{c}.$
Answer$\text{L.H.S}=\begin{vmatrix}\text{p}&\text{q}&\text{c}\\\text{a}&\text{q}&\text{c}\\\text{a}&\text{b}&\text{r} \end{vmatrix}$
$=\begin{vmatrix}\text{p}&\text{q}&\text{c}\\0&\text{q}-\text{b}&\text{c}-\text{r}\\\text{a}&\text{b}&\text{r} \end{vmatrix}$ [Applying $R_2 \rightarrow R_2 - R_3]$
$=\text{p}[\text{r}(\text{q}-\text{b})-\text{b}(\text{c}-\text{r})]+\text{a}[\text{b}(\text{c}-\text{r})-\text{c}(\text{q}-\text{b})]$
$=\text{pr}(\text{q}-\text{b})+\text{pb}(\text{r}-\text{c})-\text{ab}(\text{r}-\text{c})-\text{ac}(\text{q}-\text{b})$
$=(\text{pr}-\text{ac})(\text{q}-\text{b})+\text{b}(\text{p}-\text{a})(\text{r}-\text{c})$
Since, $\triangle=0$
$\therefore(\text{pr}-\text{ac})(\text{q}-\text{b})+\text{b}(\text{p}-\text{a})(\text{r}-\text{c})=0$
$\Rightarrow\frac{\text{pr}-\text{ac}}{(\text{p}-\text{a})(\text{r}-\text{c})}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{pr}-\text{ar}+\text{ar}-\text{ac}}{(\text{p}-\text{a})(\text{r}-\text{c})}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{r}(\text{p}-\text{a})+\text{a}(\text{r}-\text{c})}{(\text{p}-\text{a})(\text{r}-\text{c})}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{r}}{\text{r}-\text{c}}+\frac{\text{a}}{\text{p}-\text{a}}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}}\\=\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}-\frac{\text{a}}{\text{p}-\text{a}}-\frac{\text{b}}{\text{q}-\text{b}}$
$\Rightarrow\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}}=\frac{\text{p}-\text{a}}{\text{p}-\text{a}}+\frac{\text{q}-\text{b}}{\text{q}-\text{b}}$
$\Rightarrow\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}}=2$
View full question & answer→Question 905 Marks
Prove that:
$\begin{vmatrix}\frac{\text{a}^2+\text{b}^2}{\text{c}}&\text{c}&\text{c}\\\text{a}&\frac{\text{b}^2+\text{c}^2}{\text{a}}&\text{a}\\\text{b}&\text{b}&\frac{\text{c}^2+\text{a}^2}{\text{b}}\end{vmatrix}=4\text{abc}$
Answer$\text{L.H.S}=\begin{vmatrix}\frac{\text{a}^2+\text{b}^2}{\text{c}}&\text{c}&\text{c}\\\text{a}&\frac{\text{b}^2+\text{c}^2}{\text{a}}&\text{a}\\\text{b}&\text{b}&\frac{\text{c}^2+\text{a}^2}{\text{b}}\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{a}^2+\text{b}^2&\text{c}^2&\text{c}^2\\\text{a}^2&\text{b}^2+\text{c}^2&\text{a}^2\\\text{b}^2&\text{b}^2&\text{c}^2+\text{a}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{a}^2+\text{b}^2&\text{c}^2-\text{a}^2-\text{b}^2&\text{c}^2-\text{a}^2-\text{b}^2\\\text{a}^2&\text{b}^2+\text{c}^2-\text{a}^2&0\\\text{b}^2&0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}0&-2\text{b}^2&-2\text{a}^2\\\text{a}^2&\text{b}^2+\text{c}^2-\text{a}^2&0\\\text{b}^2&0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}(-\text{a}^2)\begin{vmatrix}-2\text{b}^2&-2\text{a}^2\\0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}+\text{b}^2\begin{vmatrix}-2\text{b}^2&-2\text{a}^2\\\text{b}^2+\text{c}^2-\text{a}^2&0\end{vmatrix}$
$=\frac{1}{\text{abc}}\big[-\text{a}^2\{-2\text{b}^2(\text{c}^2+\text{a}^2-\text{b}^2)\}+\text{b}^2\{0+2\text{a}^2(\text{b}^2+\text{c}^2-\text{a}^2)\}\big]$
$=\frac{1}{\text{abc}}\big[-\text{a}^2\{-2\text{b}^2\text{c}^2-2\text{b}^2\text{a}^2+2\text{b}^4\}+\text{b}^2\{2\text{a}^2\text{b}^2+2\text{a}^2\text{c}^2-2\text{a}^4\}\big]$
$=\frac{1}{\text{abc}}\big[2\text{a}^2\text{b}^2\text{c}^2+2\text{a}^4\text{b}^2-2\text{a}^4\text{b}^4+2\text{a}^2\text{b}^4+2\text{a}^2\text{b}^2\text{c}^2-2\text{a}^4\text{b}^2\big]$
$=\frac{1}{\text{abc}}4\text{a}^2\text{b}^2\text{c}^2$
$=4\text{abc}$
$=\text{R.H.S}$
View full question & answer→Question 915 Marks
Solve the following systems of linear equations by cramer's rule:
x - 2y = 4,
-3x + 5y = -7
AnswerGiven, x - 2y = 4
-3x + 5y = -7
Using the properties of determinants, we get
$\text{D}=\begin{vmatrix}1&-2\\-3&5 \end{vmatrix}=5-6=-1\neq0$
$\text{D}_1=\begin{vmatrix}4&-2\\-7&5 \end{vmatrix}=20-14=6$
$\text{D}_2=\begin{vmatrix}1&4\\-3&-7 \end{vmatrix}=-7+12=5$
Using cramer's Rule, we get
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{6}{-1}=-6$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{5}{-1}=-5$
$\therefore\text{x}=-6$ and $\text{y}=-5$
View full question & answer→Question 925 Marks
If $\text{A}=\begin{vmatrix}2&5\\2&1\end{vmatrix}$ and $\text{B}=\begin{vmatrix}4&-3\\2&5\end{vmatrix},$ verify that |AB| = |A| |B|.
AnswerLet $\text{A}=\begin{vmatrix}2&5\\2&1\end{vmatrix}$
$\Rightarrow|\text{A}|=2-10=-8$
$\text{B}=\begin{vmatrix}4&-3\\2&5\end{vmatrix}$
$\Rightarrow|\text{B}|=20+6=26$
Now $\text{AB}=\begin{vmatrix}2&5\\2&1\end{vmatrix}\begin{vmatrix}4&-3\\2&5\end{vmatrix}$
$=\begin{vmatrix}2\times4+5\times2&2\times(-3)+5\times5\\2\times4+1\times2&2\times(-3)+1\times5\end{vmatrix}$
$=\begin{vmatrix}8+10&-6+25\\8+2&-6+5\end{vmatrix}$
$=\begin{vmatrix}18&19\\10&-1\end{vmatrix}$
$\Rightarrow|\text{AB}|=18\times(-1)-(10)(19)$
$=-18-190=-208$
Now $|\text{AB}|=|\text{A}|\times|\text{B}|$
$-208=(-8)\times(26)$
$-208=-208$
Hence verified.
View full question & answer→Question 935 Marks
$\begin{vmatrix}\text{b}+\text{c}&\text{a}&\text{a}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix}=4\text{abc}$
Answer$\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{a}&\text{a}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix}$
$=\begin{vmatrix}0&-2\text{c}&-2\text{b}\\\text{b}&\text{c}+\text{a}&\text{b}\\\text{c}&\text{c}&\text{a}+\text{b}\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 - (R_2 + R_3)]$
$=\begin{vmatrix}0&-2\text{c}&-2\text{b}\\\text{b}&\text{c}+\text{a}-\text{b}&0\\\text{c}&0&\text{a}+\text{b}-\text{c}\end{vmatrix}$ [Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=0\begin{vmatrix}\text{c}+\text{a}-\text{b}&0\\0&\text{a}+\text{b}-\text{c}\end{vmatrix}-(-2\text{c})\begin{vmatrix}\text{b}&0\\\text{c}&\text{a}+\text{b}-\text{c}\end{vmatrix}-2\text{b}\begin{vmatrix}\text{b}&\text{c}+\text{a}-\text{b}\\\text{c}&0\end{vmatrix}$
$=2\text{c}[\text{b}(\text{a}+\text{b}-\text{c})-0]-2\text{b}[0-\text{c}(\text{c}+\text{a}-\text{b})]$
$=2\text{bc}[\text{a}+\text{b}-\text{c}]-2\text{bc}[\text{b}-\text{c}-\text{a}]$
$=2\text{bc}[(\text{a}+\text{b}-\text{c})-(\text{b}-\text{c}-\text{a})]$
$=4\text{abc}$
$=\text{R.H.S}$
View full question & answer→Question 945 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}0&-3&2\\0&-1&2\\0&5&2 \end{vmatrix}$ [Applying $C_1 \rightarrow C_1 + 2C_2]$
$\Rightarrow\triangle=0$
View full question & answer→Question 955 Marks
Find the real values of $\lambda$ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions:
$2\lambda\text{x}-2\text{y}+3\text{z}=0,$
$\text{x}+\lambda\text{y}+2\text{z}=0,$
$2\text{x}+0\text{y}+\lambda\text{z}=0$
AnswerThe given system of equations can be written as,
$2\lambda\text{x}-2\text{y}+3\text{z}=0$
$\text{x}+\lambda\text{y}+2\text{z}=0$
$2\text{x}+0\text{y}+\lambda\text{z}=0$
The given system of equations will have non-trivial solutions if D = 0
$\Rightarrow\begin{vmatrix}2\lambda&-2&3\\1&\lambda&2\\2&0&\lambda\end{vmatrix}=0$
$\Rightarrow 2\lambda(\lambda^2)+2(\lambda-4)+3(-2\lambda)=0$
$\Rightarrow 2\lambda^3 - 4\lambda - 8 = 0$
$\Rightarrow \lambda = 2$
So, the given system of equations will have non-trivial solutions if $\lambda = 2$
Now, we shall find solutions for $\lambda = 2$
Replacing z by k in the first two equations, we get
$2\lambda\text{x}-2\text{y}=-3\text{k}$
$\text{x}+\lambda\text{y}=-2\text{k}$
$\text{x}=\frac{\begin{vmatrix}-3\text{k}&-2\\-2\text{k}&\lambda\end{vmatrix}}{\begin{vmatrix}2\lambda&-2\\1&\lambda\end{vmatrix}}=\frac{-3\text{k}\lambda-4\text{k}}{2\lambda^2+2} $
$=\frac{-3\text{k}(2)-4\text{k}}{2(2)^2+2}=\frac{-6\text{k}-4\text{k}}{10}=-\text{k}$
$\text{y}=\frac{\begin{vmatrix}2\lambda&-3\text{k}\\1&-2\text{k}\end{vmatrix}}{\begin{vmatrix}2\lambda&-2\\1&\lambda\end{vmatrix}}=\frac{-4\text{k}\lambda+3\text{k}}{2\lambda^2+2}$
$=\frac{-4\text{k}(2)+3\text{k}}{2(2)^2+2}=\frac{-5\text{k}}{10}=\frac{-\text{k}}{2} $
Substituting these value of x and y in the third equation, we get
$\text{L.H.S}= 2(-\text{k})+0\Big(-\frac{\text{k}}{2}\Big)+2\text{k}$
$=0=\text{R.H.S}$
Thus,
$\lambda=2, \text{x}=-\text{k},\text{y}=-\frac{\text{k}}{2}$ and $\text{z}=\text{k}\ [\text{k}\in\text{R}]$
View full question & answer→Question 965 Marks
Solve the following determinant equations:
$\begin{vmatrix}3\text{x}-8&3&3\\3&3\text{x}-8&8\\3&3&3\text{x}-8\end{vmatrix}=0$
Answer$\begin{vmatrix}3\text{x}-8&3&3\\3&3\text{x}-8&8\\3&3&3\text{x}-8\end{vmatrix}=0$
Apply $C_1 \rightarrow C_1+ C_2 + C_3$
$\Rightarrow\begin{vmatrix}3\text{x}-2&3&3\\3\text{x}-2&3\text{x}-8&8\\3\text{x}-2&3&3\text{x}-8\end{vmatrix}=0$
$\Rightarrow(3\text{x}-2)\begin{vmatrix}1&3&3\\1&3\text{x}-8&8\\1&3&3\text{x}-8\end{vmatrix}=0$
$\Rightarrow(3\text{x}-2)\begin{vmatrix}1&3&3\\0&3\text{x}-11&0\\0&0&3\text{x}-11\end{vmatrix}=0$
$\Rightarrow(3\text{x}-2)(3\text{x}-11)^2=0$
$\Rightarrow(3\text{x}-2)=0$ or $(3\text{x}-11)^2=0$
$\Rightarrow\text{x}=\frac{2}{3}$ or $\text{x}=\pm\frac{11}{3}$
View full question & answer→Question 975 Marks
Solve the following systems of linear equations by cramer's rule:
2x - y = -2,
3x + 4y = 3
AnswerLet $\text{D}=\begin{vmatrix}2&-1\\3&4\end{vmatrix}=11$
$\text{D}_1=\begin{vmatrix}-2&-1\\3&4\end{vmatrix}=-5$
$\text{D}_2=\begin{vmatrix}2&-2\\3&3\end{vmatrix}=12$
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-5}{11}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{12}{11}$
View full question & answer→Question 985 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
x + y - z = 0,
x - 2y + z = 0,
3x + 6y - 5z = 0
AnswerUsing the equations we get,
$\text{D}=\begin{vmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{vmatrix}$
$=1(10-6)-1(-5-3)-1(6+6)=0$
$\text{D}_1=\begin{vmatrix}1&1&-1\\0&-2&1\\0&6&-5\end{vmatrix}$
$=0(10-6)-1(0-0)-1(0+0)=0$
$\text{D}_2=\begin{vmatrix}1&0&-1\\1&0&1\\3&0&-5\end{vmatrix}$
$=1(0-0)-0(5-3)-1(0-0)=0$
$\text{D}_3=\begin{vmatrix}1&1&0\\1&-2&0\\3&6&0\end{vmatrix}$
$=1(0-0)-1(0-0)+0(6+6)=0$
$\text{D}=\text{D}_1=\text{D}_2$
Thus, the system has infinitely many solution.
View full question & answer→Question 995 Marks
Prove that:
$\begin{vmatrix}\text{a}^2+1&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2+1&\text{bc}\\\text{ca}&\text{cb}&\text{c}^2+1 \end{vmatrix}=1+\text{a}^2+\text{b}^2+\text{c}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2+1&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2+1&\text{bc}\\\text{ca}&\text{cb}&\text{c}^2+1 \end{vmatrix}$
$=(\text{abc})\begin{vmatrix}\text{a}+\frac{1}{\text{a}}&\text{b}&\text{c}\\\text{a}&\text{b}+\frac{1}{\text{b}}&\text{c}\\\text{a}&\text{b}&\text{c}+\frac{1}{\text{c}} \end{vmatrix}$
[Taking out a, b and c common from $R_1, R_2$ and $R_3$]
$=(\text{abc})\begin{vmatrix}\text{a}+\frac{1}{\text{a}}&\text{b}&\text{c}\\-\frac{1}{\text{a}}&\frac{1}{\text{b}}&0\\-\frac{1}{\text{a}}&0&\frac{1}{\text{c}} \end{vmatrix}$
[Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1]$
$=(\text{abc})\Big(\frac{1}{\text{abc}}\Big)\begin{vmatrix}\text{a}^2+1&\text{b}^2&\text{c}^2\\-1&1&0\\-1&0&1\end{vmatrix}$
[Applying $C_1 \rightarrow aC_1, C_2 \rightarrow bC_2$ and $C_3 \rightarrow cC_3]$
$=\begin{vmatrix}\text{a}^2+1&\text{b}^2&\text{c}^2\\-1&1&0\\-1&0&1\end{vmatrix}$
$=(-1)\begin{vmatrix}\text{b}^2&\text{c}^2\\1&0\end{vmatrix}+(1)\begin{vmatrix}\text{a}^2+1&\text{b}^2\\-1&1\end{vmatrix}$
$=(-1)(-\text{c}^2)+(\text{a}^2+1+\text{b}^2)$
$=(\text{a}^2+1+\text{b}^2+\text{c}^2)$
$=(\text{a}^2+\text{b}^2+\text{c}^2+1)$
$=\text{R.H.S}$
View full question & answer→Question 1005 Marks
Evaluate the following:
$\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}$
Applying $R_1 \rightarrow R_1 - R_2$ and $R_3 \rightarrow R_3 - R_2$
$\triangle=\begin{vmatrix}\text{a}&-\text{a}&0\\\text{x}&\text{a}+\text{y}&\text{z}\\0&-\text{a}&\text{a}\end{vmatrix}$
Applying $C_2 \rightarrow C_2 + C_1$
$\triangle=\begin{vmatrix}\text{a}&0&0\\\text{x}&\text{a}+\text{y}&\text{z}\\0&-\text{a}&\text{a}\end{vmatrix}$
$\triangle=\text{a}[\text{a}(\text{a}+\text{x}+\text{y})+\text{az}]+0+0$
$\triangle=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
View full question & answer→Question 1015 Marks
Prove that:
$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z}).$
Answer$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z})$
$\text{L.H.S}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}$
$=\text{xyz}\begin{vmatrix}1&1&1\\\text{x}&\text{y}&\text{z}\\\text{x}^3&\text{y}^3&\text{z}^3\end{vmatrix}$
$=\text{xyz}\begin{vmatrix}0&1&0\\\text{x}-\text{y}&\text{y}&\text{z}-\text{y}\\\text{x}^3-\text{y}^3&\text{y}^3&\text{z}^3-\text{y}^3\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})\begin{vmatrix}0&1&0\\1&\text{y}&1\\\text{x}^2+\text{y}^2+\text{xy}&\text{y}^3&\text{z}^2+\text{y}^2+\text{zy}\end{vmatrix}$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})[\text{z}^2+\text{y}^2+\text{zy}-\text{x}^2-\text{y}^2-\text{xy}]$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})[(\text{z}-\text{x})(\text{z}+\text{x})+\text{y}(\text{z}-\text{x})]$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})(\text{z}-\text{x})[\text{z}+\text{x}+\text{y}]$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z})$
$=\text{R.H.S}$
View full question & answer→Question 1025 Marks
Prove that:
$\begin{vmatrix}1&\text{b}+\text{c}&\text{b}^2+\text{c}^2\\1&\text{c}+\text{a}&\text{c}^2+\text{a}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2 \end{vmatrix}=(\text{a}+\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$
AnswerLet $\text{L.H.S}=\begin{vmatrix}1&\text{b}+\text{c}&\text{b}^2+\text{c}^2\\1&\text{c}+\text{a}&\text{c}^2+\text{a}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2 \end{vmatrix}$
$=\begin{vmatrix}0&(\text{b}+\text{c})-(\text{c}+\text{a})&(\text{b}^2+\text{c}^2)-(\text{c}^2+\text{a}^2)\\0&(\text{c}+\text{a})-(\text{a}+\text{b})&(\text{c}^2+\text{a}^2)-(\text{a}^2+\text{b}^2)\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3]$
$=\begin{vmatrix}0&\text{b}-\text{a}&\text{b}^2-\text{a}^2\\0&\text{c}-\text{b}&\text{c}^2-\text{b}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix}$
$=(-1)^2\begin{vmatrix}0&\text{a}-\text{b}&\text{a}^2-\text{b}^2\\0&\text{b}-\text{c}&\text{b}^2-\text{c}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix}$ [Taking out (-1) common from $R_1$ and $R_2]$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}0&1&\text{a}+\text{b}\\0&1&\text{b}+\text{c}\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix}$
$=(\text{a}+\text{b})(\text{b}-\text{c})\left\{1\times\begin{vmatrix} 1 & \text{a}+\text{b} \\ 1 & \text{b}+\text{c} \end{vmatrix}\right\}$
$=(\text{a}+\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$
$=\text{R.H.S}$
View full question & answer→Question 1035 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}\text{a}&\text{h}&\text{g}\\\text{h}&\text{b}&\text{f}\\\text{g}&\text{f}&\text{c} \end{vmatrix}$
Answer$\text{M}_{11}=\begin{vmatrix}\text{b}&\text{f}\\\text{f}&\text{c} \end{vmatrix}=\text{bc}-\text{f}^2$
$\text{M}_{21}=\begin{vmatrix}\text{h}&\text{g}\\\text{f}&\text{c} \end{vmatrix}=\text{hc}-\text{fg}$
$\text{M}_{31}=\begin{vmatrix}\text{h}&\text{g}\\\text{b}&\text{f} \end{vmatrix}=\text{hf}-\text{gb}$
$\text{C}_{11}=(-1)^{1+1}\text{M}_{11}=\text{bc}-\text{f}^2$
$\text{C}_{21}=(-1)^{2+1}\text{M}_{11}=-(\text{hc}-\text{fg})=\text{fg}-\text{hc}$
$\text{C}_{31}=(-1)^{3+1}\text{M}_{31}=\text{hf}-\text{gb}$
$\text{D}=\text{a}(\text{bc}-\text{f}^2)-\text{h}(\text{hc}-\text{fg})+\text{g}(\text{fh}-\text{bg})$
$=\text{abc}-\text{af}^2-\text{h}^2\text{c}+\text{fgh}+\text{fgh}-\text{bg}^2$
$=\text{abc}+2\text{hfg}-\text{af}^2-\text{bg}^2-\text{ch}^2$
View full question & answer→Question 1045 Marks
Prove the following identities:
$\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
Answer$\text{L.H.S}=\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}$
$=\begin{vmatrix}\text{a}+\text{x}+\text{y}+\text{z}&\text{y}&\text{z}\\\text{a}+\text{x}+\text{y}+\text{z}&\text{a}+\text{y}&\text{z}\\\text{a}+\text{x}+\text{y}+\text{z}&\text{y}&\text{a}+\text{z}\end{vmatrix}$ [Applying $C_1 \rightarrow C_1 + C_2 + C_3]$
$=(\text{a}+\text{x}+\text{y}+\text{z})\begin{vmatrix}1&\text{y}&\text{z}\\1&\text{a}+\text{y}&\text{z}\\1&\text{y}&\text{a}+\text{z}\end{vmatrix}$
$=(\text{a}+\text{x}+\text{y}+\text{z})\begin{vmatrix}1&\text{y}&\text{z}\\0&\text{a}&0\\0&0&\text{a}\end{vmatrix}$ [Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_2 - R_1]$
$=(\text{a}+\text{x}+\text{y}+\text{z})\text{a}^2$ [Expanding along first column]
$=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
$=\text{R.H.S}$
$\therefore\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
View full question & answer→Question 1055 Marks
Prove that:
$\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\1&\text{b}^2+\text{ca}&\text{b}^3\\1&\text{c}^2+\text{ab}&\text{c}^3 \end{vmatrix}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}^2+\text{b}^2+\text{c}^2)$
Answer$\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\1&\text{b}^2+\text{ca}&\text{b}^3\\1&\text{c}^2+\text{ab}&\text{c}^3 \end{vmatrix}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}^2+\text{b}^2+\text{c}^2)$
$\text{L.H.S}=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\1&\text{b}^2+\text{ca}&\text{b}^3\\1&\text{c}^2+\text{ab}&\text{c}^3 \end{vmatrix}$
Apply: $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$
$=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&\text{b}^2+\text{ca}-\text{a}^2-\text{bc}&\text{b}^3-\text{a}^3\\0&\text{c}^2+\text{abb}+\text{ca}-\text{a}^2-\text{bc}&\text{c}^3-\text{a}^3 \end{vmatrix}$
$=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&(\text{b}^2-\text{a}^2)-\text{c}(\text{b}-\text{a})&\text{b}^3-\text{a}^3\\0&(\text{c}^2-\text{a}^2)-\text{b}(\text{c}-\text{a})&\text{c}^3-\text{a}^3 \end{vmatrix}$
$=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&(\text{b}-\text{a})(\text{b}+\text{a}-\text{c})&\text{b}^3-\text{a}^3\\0&(\text{c}-\text{a})(\text{c}+\text{a}-\text{b})&\text{c}^3-\text{a}^3 \end{vmatrix}$
$=(\text{b}-\text{a})(\text{c}-\text{a})\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&(\text{b}+\text{a}-\text{c})&\text{b}^2+\text{a}^2+\text{ab}\\0&(\text{c}+\text{a}-\text{b})&\text{c}^2+\text{a}^2+\text{ac} \end{vmatrix}$
$=(\text{b}-\text{a})(\text{c}-\text{a})\big[(\text{b}+\text{a}-\text{c})(\text{c}^2+\text{a}^2+\text{ac})\\-(\text{b}^2+\text{a}^2+\text{ab})(\text{c}^2+\text{a}^2+\text{ac})\big]$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$
View full question & answer→Question 1065 Marks
Solve the following determinant equations:
$\begin{vmatrix}1&1&\text{x}\\\text{p}+1&\text{p}+1&\text{p}+\text{x}\\3&\text{x}+1&\text{x}+2\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}1&1&\text{x}\\\text{p}+1&\text{p}+1&\text{p}+\text{x}\\3&\text{x}+1&\text{x}+2\end{vmatrix}=0$
$=\begin{vmatrix}1&1&\text{x}\\\text{p}&\text{p}&\text{p}\\3&\text{x}+1&\text{x}+2\end{vmatrix}$ [Applying $R_2\rightarrow R_2 - R_1]$
$=\text{p}\begin{vmatrix}1&1&\text{x}\\1&1&1\\3&\text{x}+1&\text{x}+2\end{vmatrix}$
$=\text{p}\begin{vmatrix}1&1&\text{x}\\1&1&1\\2&\text{x}&2\end{vmatrix}$
$=\text{p}\begin{vmatrix}0&1&\text{x}\\0&1&1\\2-\text{x}&\text{x}&2\end{vmatrix}$ [Applying $C_1 \rightarrow C_2 - C_1]$
$=\text{p}\left\{(2-\text{x})\times\begin{vmatrix}1&\text{x}\\1&1 \end{vmatrix}\right\}$ [Expanding along $C_1]$
$=\text{p}(2-\text{x})(1-\text{x})=0$
$\text{x}=1,2$
View full question & answer→Question 1075 Marks
Solve the following systems of linear equations by cramer's rule:
2x + 3y = 10,
x + 6y = 4
AnswerGiven, 2x + 3y = 10
x + 6y = 4
Using Cramer's Rule, we get
$\text{D}=\begin{vmatrix}2&3\\1&6\end{vmatrix}=12-3=9$
$\text{D}_1=\begin{vmatrix}10&3\\4&6\end{vmatrix}=60-12=48$
$\text{D}_2=\begin{vmatrix}2&10\\1&4\end{vmatrix}=8-10=-2$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{48}{9}=\frac{16}{3}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-2}{9}$
$\therefore\text{x}=\frac{16}{3}$ and $\text{y}=\frac{-2}{9}$
View full question & answer→Question 1085 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\frac{1}{\text{a}}&\text{a}^2&\text{bc}\\\frac{1}{\text{b}}&\text{b}^2&\text{ac}\\\frac{1}{\text{c}}&\text{c}^2&\text{ab} \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\frac{1}{\text{a}}&\text{a}^2&\text{bc}\\\frac{1}{\text{b}}&\text{b}^2&\text{ac}\\\frac{1}{\text{c}}&\text{c}^2&\text{ab} \end{vmatrix}$
$=\begin{vmatrix}1&\text{a}^3&\text{abc}\\1&\text{b}^3&\text{abc}\\1&\text{c}^3&\text{abc}\end{vmatrix}$ [Applying $R_1 \rightarrow aR_1, R_2 \rightarrow bR_2$ and $R_3 \rightarrow cR_3]$
$=\text{abc}\begin{vmatrix}1&\text{a}^3&1\\1&\text{b}^3&1\\1&\text{c}^3&1 \end{vmatrix}$
$\Rightarrow\triangle=0$
View full question & answer→Question 1095 Marks
$\begin{vmatrix}\text{b}^2+\text{c}^2&\text{ab}&\text{ac}\\\text{ba}&\text{c}^2+\text{a}^2&\text{bc}\\\text{ca}&\text{cb}&\text{a}^2+\text{b}^2\end{vmatrix}=4\text{a}^2\text{b}^2\text{c}^2$
Answer$\begin{vmatrix}\text{b}^2+\text{c}^2&\text{ab}&\text{ac}\\\text{ba}&\text{c}^2+\text{a}^2&\text{bc}\\\text{ca}&\text{cb}&\text{a}^2+\text{b}^2\end{vmatrix}=4\text{a}^2\text{b}^2\text{c}^2$
$\text{L.H.S}=\begin{vmatrix}\text{b}^2+\text{c}^2&\text{ab}&\text{ac}\\\text{ba}&\text{c}^2+\text{a}^2&\text{bc}\\\text{ca}&\text{cb}&\text{a}^2+\text{b}^2\end{vmatrix}$
Multiply $R_1, R_2$ and $R_3$ by a, b and c respectively.
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{ab}^2+\text{ac}^2&\text{a}^2\text{b}&\text{a}^2\text{c}\\\text{b}^2\text{a}&\text{bc}^2+\text{ba}^2&\text{b}^2\text{c}\\\text{c}^2\text{a}&\text{c}^2\text{b}&\text{ca}^2+\text{cb}^2\end{vmatrix}$
Take a, b, and c common from $C_1, C_2$ and $C_3$ respectively.
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}\text{b}^2+\text{c}^2&\text{a}^2&\text{a}^2\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
Now apply $R_1 \rightarrow R_1 + R_2 + R_3$
$=\begin{vmatrix}2(\text{b}^2+\text{c}^2)&2(\text{c}^2+\text{a}^2)&2(\text{a}^2+\text{b}^2)\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\begin{vmatrix}(\text{b}^2+\text{c}^2)&(\text{c}^2+\text{a}^2)&(\text{a}^2+\text{b}^2)\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\begin{vmatrix}\text{c}^2&0&\text{a}^2\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\big[\text{c}^2\{(\text{c}^2+\text{a}^2)(\text{a}^2+\text{b}^2)-\text{b}^2\text{c}^2\}+\text{a}^2\{\text{b}^2\text{c}^2-(\text{c}^2+\text{a}^2)\text{c}^2\}\big]$
$=4\text{a}^2\text{b}^2\text{c}^2$
$=\text{R.H.S}$
View full question & answer→Question 1105 Marks
Prove that:
$\begin{vmatrix}\text{b}+\text{c}&\text{a}-\text{b}&\text{a}\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}\end{vmatrix}=3\text{abc}-\text{a}^3-\text{b}^3-\text{c}^3$
Answer$\begin{vmatrix}\text{b}+\text{c}&\text{a}-\text{b}&\text{a}\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}\end{vmatrix}=3\text{abc}-\text{a}^3-\text{b}^3-\text{c}^3$
$\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{a}-\text{b}&\text{a}\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}\end{vmatrix}$
$=\begin{vmatrix}\text{b}+\text{c}+\text{a}&-\text{b}&\text{a}\\\text{c}+\text{a}+\text{b}&-\text{c}&\text{b}\\\text{a}+\text{b}+\text{c}&-\text{a}&\text{c}\end{vmatrix}$
$=-(\text{b}+\text{c}+\text{a})\begin{vmatrix}1&\text{b}&\text{a}\\1&\text{c}&\text{b}\\1&\text{a}&\text{c}\end{vmatrix}$
$=-(\text{b}+\text{c}+\text{a})\begin{vmatrix}1&\text{b}&\text{a}\\0&\text{c}-\text{b}&\text{b}-\text{a}\\0&\text{a}-\text{b}&\text{c}-\text{a}\end{vmatrix}$
$=-(\text{b}+\text{c}+\text{a})[(\text{c}-\text{b})(\text{c}-\text{a})-(\text{b}-\text{a})(\text{a}-\text{b})]$
$=3\text{abc}-\text{a}^3-\text{b}^3-\text{c}^3$
$=\text{R.H.S}$
View full question & answer→Question 1115 Marks
Show that the following systems of linear equations is inconsistent:
3x - y + 2z = 6,
2x - y + z = 2,
3x + 6y + 5z = 20
Answer$\text{D}=\begin{vmatrix}3&-1&2\\2&-1&1\\3&6&5\end{vmatrix}$
$=3(-5-6)+1(10-3)+2(12+3)=4$
Since D is non zero,
$\text{D}_1=\begin{vmatrix}6&-1&2\\2&-1&1\\20&6&5\end{vmatrix}$
$=6(-5-6)+1(10-20)+2(12+20)$
$=-66-10+64=-12$
$\text{D}_2=\begin{vmatrix}3&6&2\\2&2&1\\3&20&5\end{vmatrix}$
$=3(10-20)-6(10-3)+2(40-6)$
$=-30+42+68=-4$
$\text{D}_3=\begin{vmatrix}3&-1&6\\2&-1&2\\3&6&20\end{vmatrix}$
$=3(-20-12)+1(40-6)+6(12+3)$
$=-96+34+90=28$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-12}{4}=-3$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-4}{4}=-1$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{28}{4}=7$
$\therefore\text{x}=-3,\text{ y}=-1$ and $\text{z}=7$
View full question & answer→Question 1125 Marks
Solve the following system of homogeneous linear equations:
x + y - 2z = 0,
2x + y - 3z = 0,
5x + 4y - 9z = 0
AnswerConsider,
x + y - 2z = 0
2x + y - 9z = 0
5x + 4y - 9z = 0
$\text{D}=\begin{vmatrix}1&1&-2\\2&1&-3\\5&4&-9 \end{vmatrix}$
$=1(-9+12)-1(-18+15)-2(8-5)=0$
So, the system has infinitely many solutions, putting z = k in the first two equations,
x + y = 2k
2x + y = 3k
Using cramer's rule, we get
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{\begin{vmatrix}2\text{k}&1\\3\text{k}&1\end{vmatrix}}{\begin{vmatrix}1&1\\2&1\end{vmatrix}}=\frac{-\text{k}}{-1}=\text{k}$
$ \text{y}=\frac{\text{D}_2}{\text{D}}=\frac{\begin{vmatrix}1&2\text{k}\\1&3\text{k}\end{vmatrix}}{\begin{vmatrix}1&1\\2&1\end{vmatrix}}=\frac{-\text{k}}{-1}=\text{k}$
z = k
Clearly, these value satisfy the third equation.
Thus,
x = y = z - k $[\text{k}\in\text{R}]$
View full question & answer→Question 1135 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}0&\text{x}&\text{y}\\-\text{x}&0&\text{z}\\-\text{y}&-\text{z}&0\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}0&\text{x}&\text{y}\\-\text{x}&0&\text{z}\\-\text{y}&-\text{z}&0\end{vmatrix}$
$=\frac{\text{xyz}}{\text{xyz}}\begin{vmatrix}0&\text{x}&\text{y}\\-\text{x}&0&\text{z}\\-\text{y}&-\text{z}&0\end{vmatrix}$
$=\frac{1}{\text{xyz}}\begin{vmatrix}0&\text{xz}&\text{yz}\\-\text{xy}&0&\text{zy}\\-\text{yx}&-\text{zx}&0\end{vmatrix}$
$=\frac{1}{\text{xyz}}\begin{vmatrix}-2\text{xy}&0&2\text{yz}\\-\text{xy}&0&\text{zy}\\-\text{yx}&-\text{zx}&0\end{vmatrix}$ [Applying$ R_1 \rightarrow R_1 + R_2 + R_3]$
$=\frac{1}{\text{xyz}}\begin{vmatrix}0&0&0\\-\text{xy}&0&\text{zy}\\-\text{yx}&-\text{zx}&0\end{vmatrix}=0$ [Applying $R_1 \rightarrow R_1 - 2R_2]$
View full question & answer→Question 1145 Marks
Prove the following identities:
$\begin{vmatrix}2\text{y}&\text{y}-\text{z}-\text{x}&2\text{y}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=(\text{x}+\text{y}+\text{z})^3$
Answer$\text{L.H.S}=\begin{vmatrix}2\text{y}&\text{y}-\text{z}-\text{x}&2\text{y}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$ $[R_1 = R_1 + R_2 + R_3]$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}1&1&1\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}1&0&0\\2\text{z}&0&-\text{x}-\text{y}-\text{z}\\\text{x}-\text{y}-\text{z}&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\end{vmatrix}$ $[C_2 = C_2 - C_1, C_3 = C_3 - C_1]$
$=(\text{x}+\text{y}+\text{z})\big[1\{0+(\text{x}+\text{y}+\text{z})(\text{x}+\text{y}+\text{z})\}\big]$
$=(\text{x}+\text{y}+\text{z})^3$
$=\text{R.H.S}$
Hence proved.
View full question & answer→Question 1155 Marks
Prove the following identities:
$\begin{vmatrix}\text{a}^3&2&\text{a}\\\text{b}^3&2&\text{b}\\\text{c}^3&2&\text{c}\end{vmatrix}$
$=2(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})(\text{a}+\text{b}+\text{c})$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^3&2&\text{a}\\\text{b}^3&2&\text{b}\\\text{c}^3&2&\text{c}\end{vmatrix}$
$=2\begin{vmatrix}\text{a}^3&1&\text{a}\\\text{b}^3&1&\text{b}\\\text{c}^3&1&\text{c}\end{vmatrix}$
$=2\{\text{a}^3(\text{c}-\text{d})-1(\text{b}^3\text{c}-\text{bc}^3)+\text{a}(\text{b}^3-\text{c}^3)\}$
$=2\{\text{a}^3(\text{c}-\text{b})-\text{bc}(\text{b}-\text{c})(\text{b}+\text{c})+\text{a}(\text{b}-\text{c})(\text{b}^2+\text{bc}+\text{c}^2)\}$
$=(\text{b}-\text{c})\{-\text{a}^3-\text{bc}(\text{b}+\text{c})+\text{a}(\text{b}^2+\text{bc}+\text{c}^2)\}$
$=2(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})(\text{a}+\text{b}+\text{c})$
$=\text{R.H.S}$
View full question & answer→Question 1165 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}1^2&2^2&3^2&4^2\\2^2&3^2&4^2&5^2\\3^2&4^2&5^2&6^2\\4^2&5^2&6^2&7^2\end{vmatrix}$
Answer$\begin{vmatrix}1^2&2^2&3^2&4^2\\2^2&3^2&4^2&5^2\\3^2&4^2&5^2&6^2\\4^2&5^2&6^2&7^2\end{vmatrix}$
Applying:$ C_3 \rightarrow C_3 - C_2, C_4 \rightarrow C_4 - C_1$
$=\begin{vmatrix}1^1&2^2&3^2-2^2&4^2-1^2\\2^2&3^2&4^2-3^2&5^2-2^2\\3^3&4^2&5^2-4^2&6^2-3^2\\4^2&5^2&6^2-5^2&7^2-4^2 \end{vmatrix}$
$=\begin{vmatrix}1^1&2^2&5&15\\2^2&3^2&7&21\\3^3&4^2&9&27\\4^2&5^2&11&33 \end{vmatrix}$
Take 3 common from $C_4$
$=0$
$\because\text{C}_3=\text{C}_4$
View full question & answer→Question 1175 Marks
Evaluate the following:
$\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$\triangle=\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$=\text{x}^2\text{y}^2\text{z}^2\begin{vmatrix}0&\text{x}&\text{x}\\\text{y}&0&\text{y}\\\text{z}&\text{z}&0\end{vmatrix}$
[Taking $x ^2$ common from from $C _1, y ^2$ common from $C _2$ and $z ^2$ common from $C _3$ ]
$=\text{x}^3\text{y}^3\text{z}^3\begin{vmatrix}0&0&1\\1&-1&1\\1&1&0\end{vmatrix}$
[Applying $C_2 \rightarrow C_2 - C_3]$
$=\text{x}^3\text{y}^3\text{z}^3(1+1)$ [Expanding along first row]
$=2\text{x}^3\text{y}^3\text{z}^3$
$\therefore\ \triangle=2\text{x}^3\text{y}^3\text{z}^3$
View full question & answer→Question 1185 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sqrt{23}+\sqrt{3}&\sqrt{5}&\sqrt{5}\\\sqrt{15}+\sqrt{46}&5&\sqrt{10}\\3+\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
Answer$\begin{vmatrix}\sqrt{23}+\sqrt{3}&\sqrt{5}&\sqrt{5}\\\sqrt{15}+\sqrt{46}&5&\sqrt{10}\\3+\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
$=\begin{vmatrix}\sqrt{3}&\sqrt{5}&\sqrt{5}\\\sqrt{15}&5&\sqrt{10}\\3&\sqrt{15}&5\end{vmatrix}+\begin{vmatrix}\sqrt{23}&\sqrt{5}&\sqrt{5}\\\sqrt{46}&5&\sqrt{10}\\\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
$=\sqrt{3}\begin{vmatrix}1&\sqrt{5}&\sqrt{5}\\\sqrt{5}&5&\sqrt{10}\\\sqrt{3}&\sqrt{15}&5\end{vmatrix}+\sqrt{23}\begin{vmatrix}1&\sqrt{5}&\sqrt{5}\\\sqrt{2}&5&\sqrt{10}\\\sqrt{5}&\sqrt{15}&5\end{vmatrix}$
$=\sqrt{3}\times\sqrt{5}\begin{vmatrix}1&\sqrt{5}&\sqrt{5}\\\sqrt{5}&5&\sqrt{10}\\\sqrt{3}&\sqrt{3}&5\end{vmatrix}+\sqrt{23}\times\sqrt{5}\begin{vmatrix}1&\sqrt{5}&1\\\sqrt{2}&5&\sqrt{2}\\\sqrt{5}&\sqrt{15}&5\end{vmatrix}$
$=0+0$
$=0$
View full question & answer→Question 1195 Marks
Prove that:
$\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}=9(\text{a}+\text{b})\text{b}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$
$=\begin{vmatrix}3\text{a}+3\text{b}&3\text{a}+3\text{b}&3\text{a}+3\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$ [Applying $R_1 \rightarrow R_2 + R_2 + R_3]$
$=3(\text{a}+\text{b})\begin{vmatrix}1&1&1\\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$ [Taking out 3(a + b) common from $R_1]$
$=3(\text{a}+\text{b})\begin{vmatrix}0&0&1\\2\text{b}&-\text{b}&\text{a}+\text{b}\\-\text{b}&2\text{b}&\text{a} \end{vmatrix}$ [Applying $C_1 \rightarrow C_1 - C_2$ and $C_2 \rightarrow C_2 - C_3]$
$=3(\text{a}+\text{b})\text{b}^2\begin{vmatrix}0&0&1\\2&-1&\text{a}+\text{b}\\-1&2&\text{a} \end{vmatrix}$ [Taking out b common from $C_1$ and $C_2]$
$=3(\text{a}+\text{b})\text{b}^2\times3$
$=9(\text{a}+\text{b})\text{b}^2$
$=\text{R.H.S}$
View full question & answer→Question 1205 Marks
$\begin{vmatrix}0&\text{b}^2\text{a}&\text{c}^2\text{a}\\\text{a}^2\text{b}&0&\text{c}^2\text{b}\\\text{a}^2\text{c}&\text{b}^2\text{c}&0\end{vmatrix}=2\text{a}^3\text{b}^3\text{c}^3$
Answer$\text{L.H.S}=\begin{vmatrix}0&\text{b}^2\text{a}&\text{c}^2\text{a}\\\text{a}^2\text{b}&0&\text{c}^2\text{b}\\\text{a}^2\text{c}&\text{b}^2\text{c}&0\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}0&\text{b}^3\text{a}&\text{c}^3\text{a}\\\text{a}^3\text{b}&0&\text{c}^3\text{b}\\\text{a}^3\text{c}&\text{b}^3\text{c}&0\end{vmatrix}$ [Multiplying the three columns by a, b, and c]
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}0&\text{b}^3&\text{c}^3\\\text{a}^3&0&\text{c}^3\\\text{a}^3&\text{b}^3&0\end{vmatrix}$ [Taking out a, b and c common from the three rows]
$=\text{b}^3\begin{vmatrix}\text{b}^3&\text{c}^3\\\text{a}^3&0\end{vmatrix}+\text{c}^3\begin{vmatrix}\text{a}^3&0\\\text{a}^3&\text{b}^3\end{vmatrix}$ [Expanding along $R_1]$
$=2\text{a}^3\text{b}^3\text{c}^3$
View full question & answer→Question 1215 Marks
Show that $\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\\text{x}-2&\text{x}-3&\text{x}-\beta\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}=0,$ where $\alpha,\beta,\gamma$ are in A.P.
AnswerSince, $\alpha,\beta,\gamma$ are in A.P, $2\beta=\alpha+\gamma$
$\text{L.H.S}=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\\text{x}-2&\text{x}-3&\text{x}-\beta\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$
$\text{R}_2\rightarrow\text{R}_2-\frac{\text{R}_1}{2}-\frac{\text{R}_3}{2}$
$=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\$\text{x}-2)-\frac{\text{x}-3}{2}-\frac{\text{x}-1}{2}&(\text{x}-3)-\frac{\text{x}-4}{2}-\frac{\text{x}-2}{2}&(\text{x}-\beta)-\frac{\text{x}-\alpha}{2}-\frac{\text{x}-\gamma}{2}\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$
$=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\0&0&0\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$ $[\because2\beta=\alpha+\gamma]$
$=0$
$=\text{R.H.S}$
View full question & answer→