5 Marks Questions - Page 3 - MATHS STD 12 Science Questions - Vidyadip

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5 Marks Questions

Question 1015 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 $R_1]$
$=2\text{a}^3\text{b}^3\text{c}^3$

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Question 1025 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

Answer

Using 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.

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Question 1035 Marks

Solve the following systems of linear equations by cramer's rule:

x - 2y = 4,

-3x + 5y = -7

Answer

Given, 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 $\text{y}=-5$

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Question 1045 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$: C_2 \rightarrow C_2 - C_1$
$\begin{vmatrix}\text{a}+\text{b}&\text{a}&\text{a}\\2\text{a}+\text{b}&\text{a}&\text{a}\\4\text{a}+\text{b}&\text{a}&\text{a} \end{vmatrix}$
$=0$
$\because\text{C}_3=\text{C}_2$

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Question 1055 Marks

Prove that: $\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}=16(3\text{x}+4)$

Answer

Let $\text{L.H.S}=\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
$=\begin{vmatrix}3\text{x}+4&3\text{x}+4&3\text{x}+4\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}
[$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 $R_1]$
$=16(3\text{x}+4)$
$=\text{R.H.S}$

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Question 1065 Marks

Solve the following systems of linear equations by cramer's rule:

2x - y = 17,

3x + 5y = 6

Answer

Given, 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 $\text{y}=-3$

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Question 1075 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 $R_1 \rightarrow R_1 - R_3$
$=\begin{vmatrix}0&-2\text{x}&-2\text{x}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
$=2\text{x}[\text{z}(\text{x}+\text{y})-\text{xy}]-2\text{x}[\text{zx}-\text{y}(\text{z}+\text{x})]$
$=2\text{x}[\text{zx}+\text{zy}-\text{xy}-\text{zx}+\text{yz}+\text{yx}]$
$=4\text{xyz}$

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Question 1085 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.

Answer

Let 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.

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Question 1095 Marks

Solve the following systems of linear equations by cramer's rule $:$
$x + y + z + 1 = 0,$
$ax + by + cz + d = 0,$
$a^2x + b^2y + x^2z + d^2 = 0$

Answer

These 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,
$\text{x}=\frac{\text{D}_1}{\text{D}}=-\frac{(\text{d}-\text{b})(\text{b}-\text{c})(\text{c}-\text{d})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=-\frac{(\text{a}-\text{d})(\text{d}-\text{c})(\text{c}-\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{(\text{a}-\text{b})(\text{b}-\text{d})(\text{d}-\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$

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Question 1105 Marks

Solve the following systems of linear equations by cramer's rule: $6x + y - 3z = 5,x + 3y - 2z = 5,2x + y + 4z = 8$

Answer

Let $\text{D}=\begin{vmatrix}6&1&-3\\1&3&-2\\2&1&4\end{vmatrix}$
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, $x = 1, y = 2, z = 1$

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Question 1115 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 $\text{z}=-1$

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Question 1125 Marks

If $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2,$ find $\frac{\text{dy}}{\text{dx}}$

Answer

We have, $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin\text{ xy})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{\text{d}}{\text{dx}}(\text{x}^2)-\frac{\text{d}}{\text{dx}}(\text{y}^2)$
$\Rightarrow \cos(\text{xy})\frac{\text{d}}{\text{dx}}(\text{xy})+\Bigg\{\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}\Bigg\}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \cos(\text{xy})\Big\{\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big\}+\Bigg\{\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}(1)}{\text{x}^2}\Bigg\}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \cos(\text{xy})\Big\{\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big\}+\frac{1}{\text{x}^2}\Big(\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\Big)=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \text{x}\cos(\text{xy})\frac{\text{dy}}{\text{dx}}+\text{y}\cos(\text{xy})+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}^2}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big\{\text{x}\cos(\text{xy})+\frac{1}{\text{x}}+2\text{y}\Big\}=\frac{\text{y}}{\text{x}^2}-\text{y}\cos(\text{xy})+2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}^2\cos(\text{xy})+1+2\text{xy}}{\text{x}}\Big\}=\frac{1}{\text{x}^2}\big(\text{y}-\text{x}^2\text{y}\cos(\text{xy})+2\text{x}^2\big)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^3+\text{y}-\text{x}^2\text{y}\cos(\text{xy})}{\text{x}\big(\text{x}^2\cos(\text{xy})+1+2\text{xy}\big)}$

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Question 1135 Marks

Solve the following systems of linear equations by cramer's rule: $x + y = 5,y + z = 3,x + z = 4$

Answer

Let $\text{D}=\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}$
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$

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Question 1145 Marks

Solve the following systems of linear equations by cramer's rule:

x + y = 1,

x + z = -6,

x - y - 2z = 3

Answer

These 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 $\text{z}=-4$

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Question 1155 Marks

Solve the following systems of linear equations by cramer's rule:

$x + y + z + w = 2,$

$x - 2y + 2z + 2w = -6,$

$2x + y - 2z + 2w = -5,$

$3x - y + 3z - 3w = -3$

Answer

Here, $\text{D}=\begin{vmatrix}1&1&1&1\\1&-2&2&2\\2&1&-2&2\\3&-1&3&-3\end{vmatrix}$
$\therefore\text{D}=\begin{vmatrix}1&0&0&0\\1&-3&1&1\\2&-1&-4&0\\3&-4&0&-6\end{vmatrix}=1\begin{vmatrix}-3&1&1\\-1&-4&0\\-4&0&-6\end{vmatrix}$
$[C_2 \rightarrow C_2 - C_1, C_3 \rightarrow C_3 - C_1, C_4 \rightarrow C_4 - C_1]$
$\therefore\text{D}=\begin{vmatrix}0&0&1\\-1&-4&0\\-22&6&-6\end{vmatrix} [C_1 \rightarrow C_1 + 3C_3, C_2 \rightarrow C_2 - C_3]$
$=1(-6-88)=-94$
$\text{D}_1=\begin{vmatrix}2&1&1&1\\-6&-2&2&2\\-5&1&-2&2\\-3&-1&3&-3\end{vmatrix}=188$
$\text{D}_2=\begin{vmatrix}1&2&1&1\\1&-6&2&2\\2&-5&-2&2\\3&-3&3&-3\end{vmatrix}=-282$
$\text{D}_3=\begin{vmatrix}1&1&2&1\\1&-2&-6&2\\2&-1&-5&-2\\3&-1&-3&-3\end{vmatrix}=-141$
$\text{D}_4=\begin{vmatrix}1&1&1&2\\1&-2&2&-6\\2&1&-2&-5\\3&-1&3&-3\end{vmatrix}=47$
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, $\text{x}=-2,\text{y}=3,\text{z}=\frac{3}{2},\text{w}=-\frac{1}{2}$

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Question 1165 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

Answer

We 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.
$\text{x}=\frac{5}{3},\text{ y}=\frac{-4+3\text{k}}{3},\text{ z}=\text{ k}$

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Question 1175 Marks

Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}0&2&6\\1&5&0\\3&7&1 \end{vmatrix}$

Answer

Let $M_{ij}$ 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.
$|\text{A}|=\text{a}_{11}\text{C}_{11}+\text{a}_{21}\text{C}_{21}+\text{a}_{31}\text{C}_{31}$
$=0\times5+1\times(40)+3\times(-30)$
$=40-90$
$=-50$

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Question 1185 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$: R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - R_1$
$=(\text{b}+\text{c}+\text{a})\begin{vmatrix}\text{a}&1&\text{a}^2\\\text{b}-\text{a}&0&\text{b}^2-\text{a}^2\\\text{c}-\text{a}&0&\text{c}^2-\text{a}^2\end{vmatrix}$
$=(\text{b}+\text{c}+\text{a})(\text{b}-\text{a})(\text{c}-\text{a})\begin{vmatrix}\text{a}&1&\text{a}^2\\1&0&\text{b}+\text{a}\\1&0&\text{c}+\text{a}\end{vmatrix}$
$=(\text{b}+\text{c}+\text{a})(\text{b}-\text{a})(\text{c}-\text{a})(\text{b}-\text{c})$

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Question 1195 Marks

Solve the following system of homogeneous linear equations:

x + y - 2z = 0,

2x + y - 3z = 0,

5x + 4y - 9z = 0

Answer

Consider,
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 $[\text{k}\in\text{R}]$

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Question 1205 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 $R_1 \rightarrow R_1 + R_2 + R_3$
$=\begin{vmatrix}2(\text{b}^2+\text{c}^2)&2(\text{c}^2+\text{a}^2)&2(\text{a}^2+\text{b}^2)\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\begin{vmatrix}(\text{b}^2+\text{c}^2)&(\text{c}^2+\text{a}^2)&(\text{a}^2+\text{b}^2)\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\begin{vmatrix}\text{c}^2&0&\text{a}^2\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\big[\text{c}^2\{(\text{c}^2+\text{a}^2)(\text{a}^2+\text{b}^2)-\text{b}^2\text{c}^2\}+\text{a}^2\{\text{b}^2\text{c}^2-(\text{c}^2+\text{a}^2)\text{c}^2\}\big]$
$=4\text{a}^2\text{b}^2\text{c}^2$
$=\text{R.H.S}$

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Question 1215 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 $R_1 \rightarrow R_1 - 2R_2]$

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5 Marks Questions - Page 3 - MATHS STD 12 Science Questions - Vidyadip