Question 1015 Marks
Using Cofactors of elements of third column, evaluate $\triangle=\begin{vmatrix}1&x&yz\\1&y&zx\\1&z&xy\end{vmatrix}.$
AnswerThe given determinant is $\begin{vmatrix}1&x&yz\\1&y&zx\\1&z&xy\end{vmatrix}.$
We have:
$M_{13} = \begin{vmatrix}1&y\\1&z\end{vmatrix}=z-y$
$M_{23} = \begin{vmatrix}1&x\\1&z\end{vmatrix}=z-x$
$M_{33} = \begin{vmatrix}1&x\\1&y\end{vmatrix}=y-x$
$\therefore A_{13} =$ cofactor of $a_{13} = (-1)^{1+3} M_{13} = (z - y)$
$A_{23} =$ cofactors of $a_{23} = (-1)^{2+3 }M_{23} = - (z - x) = (x - z)$
$A_{33} =$ cofactors of $a_{33} = (-1)^{3+3 }M_{33} = (y - x)$
We know that $\triangle$ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
$\therefore\triangle = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}$
$= yz(z - y) + zx (x - z) + xy (y - x)$
$= yz^2 - y^2z + x^2z - xz^2 + xy^2 - x^2y$
$=(x^2z - y^2z) + (yz^2 - xz^2) + (xy^2 - x^2y)$
$= z(x^2 - y^2) + z^2(y - x) + xy(y - x)$
$=z(x - y)(x + y) + z^2(y - x) + xy(y - x)$
$=(x- y)[zx + zy - z^2 - xy]$
$=(x - y)[z(x - z) + y(z - x)]$
$=(x - y)(z - x)[-z + y]$
$=(x - y)(y - z)(z - x)$
View full question & answer→Question 1025 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 1035 Marks
Show that the following system of linear equations is consistent and also find solution:
$2x + 3y = 5$
$6x + 9y = 15$
Answer$2\text{x}+3\text{y}=5\dots(1)$
$6\text{x}+9\text{y}=15\dots(2)$
Or , $AX = B$
Where,
$\text{A}=\begin{bmatrix}2&3\\ 6&9\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}5\\ 15\end{bmatrix}$
$\begin{bmatrix}2&3\\ 6&9\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}5\\ 15\end{bmatrix}$
$\therefore\ |\text{A}|=\begin{vmatrix}2&3\\6&9\end{vmatrix}$
$=18-18$
$=0$
So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because $(\text{adj A})\text{B}\neq0\text{ or }(\text{adj A})=0$.
$C_{11} = 9, C_{12} = -6, C_{21} = -3$ and $C_{22} = 2$
$\therefore\ \text{adj A}=\begin{bmatrix}9&-6\\-3&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}2&-3\\-6&9\end{bmatrix}$
$\Rightarrow(\text{adj A})\text{B}=\begin{bmatrix}9&-3\\-6&2\end{bmatrix}\begin{bmatrix}5\\15\end{bmatrix}$
$=\begin{bmatrix}45-45\\-30+30\end{bmatrix}$
$=\begin{bmatrix}0\\0\end{bmatrix}$
If $|A| = 0$ and $(adj\ A) B = 0,$ then the system is consistent and has infinitely many solutions.
Thus, $AX = B$ has infinitely many solutions.
Substituting $y = k$ in eq. $(1),$ we get
$2\text{x} + 3\text{k}=5$
$\Rightarrow2\text{x}=5-3\text{k}$
$\Rightarrow\text{x}=\frac{5-3\text{k}}{2}\text{ and }\text{y}=\text{k}$
These values of $x$ and $y$ satisfy the third equation.
Thus, $\Rightarrow\text{x}=\frac{5-3\text{k}}{2}\text{ and }\text{y}=\text{k} ($where $k$ is a real number$)$ satisfy the given system of equations.
View full question & answer→Question 1045 Marks
Using properties of determinants, prove that:
$\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\\beta&\beta^2&\gamma+\alpha\\\gamma&\gamma^2&\alpha+\beta\end{vmatrix}=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$
Answer$\triangle=\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\\beta&\beta^2&\gamma+\alpha\\\gamma&\gamma^2&\alpha+\beta\end{vmatrix}$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_{3 }\rightarrow R_{3 }- R_1,$ we have:
$\triangle=\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\\beta-\alpha&\beta^2-\alpha^2&\alpha-\beta\\\gamma-\alpha&\gamma^2-\alpha^2&\alpha-\gamma\end{vmatrix}$
$=(\beta-\alpha)(\gamma-\alpha)\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\1&\beta+\alpha&-1\\0&\gamma+\alpha&-1\end{vmatrix}$
Applying $R_3 \rightarrow R_3 - R_2,$ we have:
$\triangle=(\beta-\alpha)(\gamma-\alpha)\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\1&\beta+\alpha&-1\\1&\gamma-\beta&0\end{vmatrix}$
Expanding along $R_3,$ we have:
$\triangle=(\beta-\alpha)(\gamma-\alpha)[-(\gamma-\beta)(-\alpha-\beta-\gamma)]$
$=(\beta-\alpha)(\gamma-\alpha)(\gamma-\beta)(\alpha+\beta+\gamma)$
$=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)$
Hence, the given result is proved.
View full question & answer→Question 1055 Marks
Show that $\triangle\text{ABC}$ is an isosceles triangle, if the determinant $\triangle=\begin{vmatrix}1&1&1\\1+\cos\text{A}&1+\cos\text{B}&1+\cos\text{C}\\\cos^2\text{A}+\cos\text{A}&\cos^2\text{B}+\cos\text{B}&\cos^2\text{C}+\cos\text{C}\end{vmatrix}=0.$
AnswerWe have, $\triangle=\begin{vmatrix}1&1&1\\1+\cos\text{A}&1+\cos\text{B}&1+\cos\text{C}\\\cos^2\text{A}+\cos\text{A}&\cos^2\text{B}+\cos\text{B}&\cos^2\text{C}+\cos\text{C}\end{vmatrix}=0$
$ \triangle=\begin{vmatrix}0&0&0\\\cos\text{A}-\cos\text{C}&\cos\text{B}-\cos\text{C}&1+\cos\text{C}\\\cos^2\text{A}+\cos\text{A}-\cos^2\text{C}-\cos\text{C}&\cos^2\text{B}+\cos\text{B}-\cos^2\text{C}-\cos\text{C}&\cos^2\text{C}+\cos\text{C}\end{vmatrix}=0.$
$[\text{C}_1\rightarrow\text{C}_1-\text{C}_3\text{ and C}_2\rightarrow\text{C}_2+\text{C}_3]$
$=\begin{vmatrix}0&0&1\\1 & 1&1+\cos\text{C}\\\cos\text{A}+\cos\text{C}+1&\cos\text{B}+\cos\text{C}+1&\cos^2\text{C}+\cos\text{C}\end{vmatrix}=0$
$\big[\text{Taking }(\cos\text{A}\cos\text{C})\text{ common from C}_1\text{ and }(\cos\text{B}-\cos\text{C})\text{ common from C}_2\big]$
$\Rightarrow\ (\cos\text{A}-\cos\text{C}).(\cos\text{B}-\cos\text{C})\big[(\cos\text{B}+\cos\text{C}+1)-(\cos\text{A}+\cos\text{C}+1)\big]=0$
$\Rightarrow\ (\cos\text{A}-\cos\text{C}).(\cos\text{B}-\cos\text{C})(\cos\text{B}+\cos\text{C}+1-\cos\text{A}-\cos\text{C}-1)=0$
$\Rightarrow\ (\cos\text{A}-\cos\text{C}).(\cos\text{B}-\cos\text{C})(\cos\text{B}-\cos\text{A})=0$
$\text{i.e., }\cos\text{A}=\cos\text{C or }\cos\text{B}=\cos\text{C or}\cos\text{B}=\cos\text{A}$
$\Rightarrow\ \text{A = C or B = C or B = A}$
Hence, ABC is an isosceles triangle.
View full question & answer→Question 1065 Marks
If $\begin{bmatrix}4-\text{x}&4+\text{x}&4+\text{x}\\4+\text{x}&4-\text{x}&4+\text{x}\\4+\text{x}&4+\text{x}&4-\text{x}\end{bmatrix}=0,$ then find values of $x$.
AnswerGiven, $\begin{bmatrix}4-\text{x}&4+\text{x}&4+\text{x}\\4+\text{x}&4-\text{x}&4+\text{x}\\4+\text{x}&4+\text{x}&4-\text{x}\end{bmatrix}=0$
$\Rightarrow\ \begin{bmatrix}12+\text{x}&12+\text{x}&12+\text{x}\\4+\text{x}&4-\text{x}&4+\text{x}\\4+\text{x}&4+\text{x}&4-\text{x}\end{bmatrix}=0$ $[\because\ \text{R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3]$
$\Rightarrow\ (12+\text{x})\begin{bmatrix}1&1&1\\4+\text{x}&4-\text{x}&4+\text{x}\\4+\text{x}&4+\text{x}&4-\text{x}\end{bmatrix}=0\ [$Taking $(12 + x)$ common from $R_1]$
$\Rightarrow\ (12+\text{x})\begin{vmatrix}0&0&0\\0&8&4+\text{x}\\2\text{x}&8&4-\text{x}\end{vmatrix}=0$ $\big[\because\text{C}_1\rightarrow\text{C}_ 1-\text{C}_ 3\text{ and }\text{C}_ 2\rightarrow\text{C}_ 2+\text{C}_ 3]$
$\Rightarrow\ (12+\text{x})\big[1.(-16\text{x})\big]=0$
$\Rightarrow\ (12+\text{x})(-16\text{x})=0$
$\therefore\ \text{x}=-12,0$
View full question & answer→Question 1075 Marks
Solve the following systems of homogeneous linear equations by matrix method:
2x - y + z = 0
3x + 2y - z = 0
x + 4y + 3z = 0
AnswerThe given system of homogeneous equations can be written in matrix form as follows:
$\begin{bmatrix}2&-1&1\\3&2&-1\\1&4&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
or, $\text{AX}=\text{O}$
where, $\text{A}=\begin{bmatrix}2&-1&1\\3&2&-1\\1&4&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
$|\text{A}|=\begin{vmatrix}2&-1&1\\3&2&-1\\1&4&3\end{vmatrix}$
$=2(6+4)+1(9+1)+1(12-2)$
$=40$
$\therefore\ |\text{A}|\neq0$
So, the given system has only trivial solution, which is given below:
x = y = z = 0
View full question & answer→Question 1085 Marks
Show that the following system of linear equation is inconsistent:
$2x + 3y = 5$
$6x + 9y = 10$
AnswerThe given system of equations can be expresesed as follows:
$\text{AX = B}$
Here,
$\text{A}=\begin{bmatrix}2&3\\ 6&9\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}5\\ 10\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}2&3\\ 6&9\end{vmatrix}$
$={(18-18)}$
$=0$
Let $C_{ij}$ be the co-factors of the elements $a_{ij}$ in $A = [a_{ij}].$ Then,
$\text{C}_{11}={(-1)}^{1+1}{(9)}=9,\\ \text{C}_{12}={(-1)}^{1+2}{(6)}=-6$
$\text{C}_{21}={(-1)}^{2+1}{(3)}=-3,\\ \text{C}_{22}={(-1)}^{2+2}{(6)}=2$
$\text{adj A}=\begin{bmatrix}9&-6\\ -3&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}9&-3\\ -6&2\end{bmatrix}$
$\text{(adj A) = B}=\begin{bmatrix}9&-3\\ -6&2\end{bmatrix}\begin{bmatrix}5\\ 10\end{bmatrix}$
$=\begin{bmatrix}45-30\\ -30+30\end{bmatrix}$
$=\begin{bmatrix}15\\ -10\end{bmatrix}\neq0$
Hence, the given system of equations is inconsistent.
View full question & answer→Question 1095 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 1105 Marks
If $\text{A}=\begin{bmatrix}2 & 3 \\ 1 & 2 \end{bmatrix},$ verify that $A^2 - 4A + I = 0,$ where $\text{I}=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\text{ and O}\begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix}.$ Hence find $A^{-1}.$
Answer$\text{A}=\begin{bmatrix}2 & 3 \\ 1 & 2 \end{bmatrix}$
$\therefore\ \text{A}^2=\begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}$
and $\text{A}^2-4\text{A}+\text{I}$
$=\begin{bmatrix}7 & 12 \\4 & 7 \end{bmatrix}-\begin{bmatrix}8 & 12 \\4 & 8 \end{bmatrix}+\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}$
$\Rightarrow\ \text{A}^2-4\text{A}+\text{I}$
$=\begin{bmatrix}7-8+1 & 12-12+0 \\4-4+0 & 7-8+1 \end{bmatrix}$
$=\begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix}=0$
$\Rightarrow A^2 - 4A + I$
$\Rightarrow A^2 - 4A = -I$
$\Rightarrow A^{-1}A^2 - 4AA^{-1} = -IA^{-1} [$Pre-multiplying both sides by $A^{-1}]$
$\Rightarrow A - 4I = -A^{-1}$
$\Rightarrow A^{-1} = 4I - A$
$\Rightarrow\ \text{A}^{-1}=\left\{\begin{bmatrix}4 & 0 \\0 & 4 \end{bmatrix}-\begin{bmatrix}2 & 3 \\ 1 & 2 \end{bmatrix}\right\}$
$=\begin{bmatrix}2 & -3 \\-1 & 2 \end{bmatrix}$
View full question & answer→Question 1115 Marks
The cost of $4 \ kg$ onion, $3 \ kg$ wheat and $2 \ kg$ rice is $Rs .60.$ The cost of $2 \ kg$ onion, $4 \ kg$ wheat and $6 \ kg$ rice is $Rs 90.$ The cost of $6 \ kg$ onion $2 \ kg$ wheat and $3 \ kg$ rice is $Rs .70.$ Find cost of each item per $kg$ by matrix method.
AnswerLet $₹. x, ₹. y, ₹ .z$ per $kg$ be the prices of onion, wheat and rice respectively.
$\therefore$ According to given data,
we have three equations,$4x + 3y + 2z = 60,2x + 4y + 6z = 90,6x + 2y + 3z = 70$
Matrix form of given equations is $AX = B \Rightarrow\ \begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}60\\90\\70\end{bmatrix}$ $\text{Here}\ \text{A}=\begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{and B}=\begin{bmatrix}60\\90\\70\end{bmatrix}$
$\therefore\ \left|\text{A}\right|=\begin{vmatrix}4&3&2\\2&4&6\\6&2&3\end{vmatrix}$
$=4(12-12)-3(6-36)+2(4-24)$
$=0+90-40$
$=50\neq0$
Therefore, solution is unique and $\text{X=A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(adj. A)B} \dots\dots(1)$
Now, $A_{11} = 0, A_{12} = 30, A_{13} = -20 A_{21} = -5, A_{22} = 0, A_{23} = 10 ,A_{31} = 10, A_{32} = -20,_{ }A_{33} = 10$
$\therefore\ \text{adj.A}=\begin{bmatrix}0&30&-20\\-5&0&10\\10&-20&10\end{bmatrix}=\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}$
$\Rightarrow$ From $eq. (1), \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}\begin{bmatrix}60\\90\\70\end{bmatrix}$
$=\frac{1}{50}\begin{bmatrix}-450+700\\1800-1400\\-1200+900+700\end{bmatrix}$
$=\frac{1}{50}\begin{bmatrix}250\\400\\400\end{bmatrix}$
$=\begin{bmatrix}5\\8\\8\end{bmatrix}$
Therefore, $x = 5, y = 8$ and $z = 8$
Hence, the cost of onion, wheat and rice are $₹ .5, ₹ .8$ and $₹ .8$ per $kg.$
View full question & answer→Question 1125 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 1135 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 1145 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 1155 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 1165 Marks
Let $\text{A}=\begin{bmatrix}3 & 2 \\7 & 5 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}6 & 7 \\8 & 9 \end{bmatrix}$. Find $(AB)^{-1}.$
Answer$\text{A}=\begin{bmatrix}3 & 2\\7 & 5 \end{bmatrix}\therefore\ |\text{A}|=1\neq0$ and $\text{adj A}=\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$
$\therefore\ \text{A}^{-1}\frac{\text{adj A}}{|\text{A}|}=\frac{1}{1}\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$
$\text{B}=\begin{bmatrix}6 & 7 \\7 & 9 \end{bmatrix}\therefore\ |\text{B}|=-2\neq0$ and $\text{adj B}=\begin{bmatrix}9 & -7 \\-8 & 6 \end{bmatrix}$
$\therefore\ \text{B}^{-1}=\frac{\text{adj B}}{|\text{B}|}=\frac{1}{(-2)}=\begin{bmatrix}9 & -7 \\-8 & 6 \end{bmatrix}$
Now, $(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1}$
$(\text{AB})^{-1}=\frac{1}{(-2)}\begin{bmatrix}9 & -7 \\-8 & 6 \end{bmatrix}\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$
$(\text{AB})^{-1}=-\frac{1}{2}\begin{bmatrix}94 & -39 \\-82 & 34 \end{bmatrix}$
$\text{(AB)}^{-1}=\begin{bmatrix}-47 & \frac{39}{2} \\41 & -17 \end{bmatrix}$
View full question & answer→Question 1175 Marks
If $\text{A}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix},$ find the value of $\lambda$ so that $\text{A}^2=\lambda\text{A}-2\text{I}.$ Hence, find $A^{-1}.$
Answer$\text{A}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}$
$\text{A}^2=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}$
$=\begin{bmatrix}1 & -2 \\4 & -4 \end{bmatrix}$
If $\text{A}^2=\lambda\text{A}-2\text{I}$
$\lambda\text{A}=\text{A}^2+2\text{I}$
$=\begin{bmatrix}1 & -2 \\4 & -4 \end{bmatrix}+\begin{bmatrix}2 & 0 \\0 & 2 \end{bmatrix}$
$\lambda\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}$
$\lambda\begin{bmatrix}3\lambda & -2\lambda \\4\lambda & -2\lambda \end{bmatrix}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}$
$3\lambda=3$
$\lambda=1$
$\text{A}^2=\text{A}-2\text{I}$
$Px$ multiplying by $A^{-1}$
$A^{-1} AA = A^{-1} A - A^{-1} I$
$A = I - 2A^{-1}$
$2\text{A}^{-1}=\text{I}-\text{A}=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}-\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}=\begin{bmatrix}-2 & 2 \\-4 & 3 \end{bmatrix}$
$\therefore\ \text{A}^{-1}=\frac{1}{2}\begin{bmatrix}-2 & 2 \\-4 & 3 \end{bmatrix}$
View full question & answer→Question 1185 Marks
Find the matrix $X$ satisfying the equation:
$\begin{bmatrix}2 & 1 \\5 & 3 \end{bmatrix}\text{X}\begin{bmatrix}5 & 3 \\3 & 2 \end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}.$
AnswerLet $\text{A}=\begin{bmatrix}2 & 1 \\5 & 3 \end{bmatrix}$
$\text{B}=\begin{bmatrix}5 & 3 \\3 & 2 \end{bmatrix}$
Then the given equation can be wirtten as
$A \times B = I$
$\Rightarrow X = A^{-1}B^{-1}$
Now $|A| = 6 - 5 = 1$
$|B| = 10 - 9 = 1$
$\text{A}^{-1}=\frac{\text{adj (A)}}{|\text{A}|}=\begin{bmatrix}3 & -1 \\-5 & 2 \end{bmatrix}$
$\text{B}^{-1}=\frac{\text{adj (B)}}{|\text{B}|}=\begin{bmatrix}2 & -3 \\-3 & 5 \end{bmatrix}$
$\therefore\ \text{X}=\begin{bmatrix} 3 & -1 \\-5 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\-3 & 2 \end{bmatrix}$
$=\begin{bmatrix} 9 & -14 \\-16 & 25 \end{bmatrix}$
View full question & answer→Question 1195 Marks
Prove that:
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\\text{a}+2)(\text{a}+3)&\text{a}+3&1\\\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}=-2$
Answer$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\\text{a}+2)(\text{a}+3)&\text{a}+3&1\\\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}=-2$
$\text{L.H.S}=\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\\text{a}+2)(\text{a}+3)&\text{a}+3&1\\\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}$
Apply $R_3 \rightarrow R_3 - R_2$
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\\text{a}+2)(\text{a}+3)&\text{a}+3&1\\\text{a}+3)2&1&0\end{vmatrix}$
Apply $R_2 \rightarrow R_2 - R_1$
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\\text{a}+2)2&1&0\\\text{a}+3)2&1&0\end{vmatrix}$
$=[(2\text{a}+4)(1)-(1)(2\text{a}+6)]$
$=-2$
$=\text{R.H.S}$
View full question & answer→Question 1205 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 1215 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 1225 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 1235 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 1245 Marks
Find the adjoint of the matrix $\text{A}=\begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$ and hence show that $\ce{A (adj A) = |A|I_3.}$
Answer$\text{A}=\begin{bmatrix} -9 & 8 &-2 \\ -8 & 7 & 2 \\ -5 & 4 & -1 \end{bmatrix}$
Now, to find $\ce{Adj A}$
$\ce{A_{11} = (-1)^{1+1} (-3) = -3, A_{21} = (-1)^{2+1} (-6) = 6, A_{31} = (-1)^{3+1} (6) = 6}$
$\ce{A_{12} = (-1)^{1+2} (6) = -6, A_{22} = (-1)^{2+2} (3) = 3, A_{32} = (-1)^{3+2} (6) = -6}$
$\ce{A_{13} = (-1)^{1+3} (-6) = -6, A_{23} = (-1)^{2+3} (6) = -6, A_{33} = (-1)^{3+3} (3) = 3}$
Therefore,
$\text{Adj .A}=\begin{bmatrix} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}$
$|\text{A}|=\begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$
$= -1(1 - 4) - 2(-2 - 4) + 2(4 + 2)$
$= 3 + 12 + 12$
$= 27$
To show: $\ce{A(adj A) = |A|I_3}$
$\text{L.H.S}=\begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}\begin{bmatrix} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}$
$\text{L.H.S}=\begin{bmatrix} 3+12+12 & -6-6+12 & -6+12-6 \\ -6-6+12 & 12+3+12 & 12-6-6 \\ -6+12-6 & 12-6-6 & 12+12+3 \end{bmatrix}$
$=\begin{bmatrix} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{bmatrix}$
$=27\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=|\text{A}|\text{I}_3$
$=\text{R.H.S}$
Hence, $\ce{A(adj A) = |A|I_3.}$
View full question & answer→Question 1255 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 1265 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 1275 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 1285 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 1295 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 1305 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 1315 Marks
Given $\text{A}=\begin{bmatrix}5 & 0 & 4 \\2 & 3 & 2 \\ 1 & 2 & 1\end{bmatrix},\text{B}^{-1}=\begin{bmatrix}1 & 3 & 3 \\1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}.$ Compute $(AB)^{-1}.$
AnswerWe have,
$\text{A}=\begin{bmatrix}5 & 0 & 4 \\2 & 3 & 2 \\ 1 & 2 & 1\end{bmatrix}$
$\text{B}^{-1}=\begin{bmatrix}1 & 3 & 3 \\1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}$
We have $(AB)^{-1} = B^{-1}A^{-1}$
For matrix $A,$
$\text{C}_{11}=\begin{vmatrix}3 & 2 \\2 & 1 \end{vmatrix}=-1,\ \text{C}_{12}=-\begin{vmatrix}2 & 2 \\1 & 1 \end{vmatrix}=0$
and $\text{C}_{13}=\begin{vmatrix}2 & 3 \\1 & 2 \end{vmatrix}=1$
$\text{C}_{21}=-\begin{vmatrix}0 & 4 \\2 & 1 \end{vmatrix}=8,\ \text{C}_{22}=\begin{vmatrix}5 & 4 \\1 & 1 \end{vmatrix}=1$
and $\text{C}_{23}=-\begin{vmatrix}5 & 0 \\1 & 2 \end{vmatrix}=-10$
$\text{C}_{31}=\begin{vmatrix}0 & 4 \\3 & 1 \end{vmatrix}=-12,\ \text{C}_{32}=-\begin{vmatrix}5 & 4 \\2 & 2 \end{vmatrix}=-2$
and $\text{C}_{33}=\begin{vmatrix}5 & 0 \\2 & 3 \end{vmatrix}=15$
Now, $\text{adj (A)}=\begin{bmatrix}-1 & 0 & 1 \\8 & 1 & -2 \\ -12 & -2 & 15 \end{bmatrix}^\text{T}$
$=\begin{bmatrix} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{bmatrix}$
and $|A| = -1$
$\therefore\text{A}^{-1}=-\begin{bmatrix} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{bmatrix}$
$=\begin{bmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{bmatrix}$
So, $\text{B}^{-1}\text{A}^{-1}=\begin{bmatrix} 1 & 3 & 3 \\1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{bmatrix}$
$=\begin{bmatrix} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{bmatrix}$
View full question & answer→Question 1325 Marks
If $\text{A}=\begin{bmatrix}-1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix},$ show that $A^2 = A^{-1}.$
AnswerWe have $\text{A}=\begin{bmatrix}-1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
Now,
$\text{A}^2=\begin{bmatrix}-1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix}-1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
$=\begin{bmatrix}1-2+0 & -2+2+0 & 0+2+0 \\ 1-1+0 & -2+1+1 & 0+1+0 \\ 0-1+0 & 0+1+0 & 0+1+0 \end{bmatrix}$
$=\begin{bmatrix}-1 & 0 & 2 \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{bmatrix}$
And $\text{A}^2\times\text{A}=\begin{bmatrix}-1 & 0 & 2 \\ 0 & 0 & 1 \\ -1 & 1 & 1\end{bmatrix}\begin{bmatrix}-1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
$=\begin{bmatrix}1+0+0 & -2+0+2 & 0 \\ 0+0+0 & 0+0+1 & 0 \\ 0 & -2+1+1 & 1 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\text{I}_3\ \big[$Identity matrix of order $3\big]$
$\Rightarrow\ \text{A}^2\times\text{A}=\text{I}_3$
$\Rightarrow\ \text{A}^2=\text{A}^{-1}$
View full question & answer→Question 1335 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 1345 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 1355 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 1365 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 1375 Marks
Solve the following system of equations by matrix method:
$5x + 7y + 2 = 0$
$4x + 6y + 3 = 0$
AnswerThe given system of equations can be written in matrix from as follws: $\begin{bmatrix}5&7\\ 4&6\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-2\\ -3\end{bmatrix}$ $\text{AX = B}$
Here, $\text{A}=\begin{bmatrix}5&7\\ 4&6\end{bmatrix},\text{X=}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}$ And$\text{ B}=\begin{bmatrix}-2\\ -3\end{bmatrix}$
Now, $|\text{A}|=\begin{bmatrix}5&7\\ 4&6\end{bmatrix}\\ $ $=30-28$ $=2\neq0$ The given system has a unique solution given by $\text{X}=\text{A}^{-1 }\text{B.}$
Let $C_{ij}$ be the cofactors of the elements $a_{ij}$ in $A = [a_{ij}]. $
Then, $\text{C}_{11}=(-1)^{1+1}(6)=6,\text{C}_{12}=(-1)^{1+2}(4)=-4$
$\text{C}_{21}=(-1)^{2+1}(7)=-7,\text{C}_{22}=(-1)^{2+2}(5)$
$\text{adj}\ \text{A}=\begin{bmatrix}6&-4\\ -7&5\end{bmatrix}=\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}$ $\text{A}^{-1}=\frac{1}{\text{|A|}}\text{ adj}\text{ A}$
$\Rightarrow \text{A}^{-1}=\frac{1}{2}\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}$ $\text{X}=\text{A}^{-1}\text{B}$
$=\frac{1}{2}\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}\begin{bmatrix}-2\\-3\end{bmatrix} $
$=\frac{1}{2}\begin{bmatrix}-12+21\\ 8-15\end{bmatrix} $
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{9}{2}\\ \frac{-7}{2}\end{bmatrix}$ $\therefore\text{X}=\frac{9}{2}$ And $\text{ y }=\frac{-7}{2}$
View full question & answer→Question 1385 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 1395 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 1405 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 1415 Marks
For the matrix $\text{A} = \begin{bmatrix}3&2\\1&1\end{bmatrix},$ find the numbers $a$ and $b$ such that $A^2 + aA + bI = 0.$
Answer$\text{Given:}\ \text{A}=\begin{bmatrix}3&2\\1&1\end{bmatrix}$ $\therefore\ \text{A}^2=\text{A.A}=\begin{bmatrix}3&2\\1&1\end{bmatrix}\begin{bmatrix}3&2\\1&1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}9+2&6+2\\3+1&2+1\end{bmatrix}=\begin{bmatrix}11&8\\4&3\end{bmatrix}$
$\therefore\ \text{A}^2+\text{aA}+\text{bI}=0$
$\Rightarrow\ \begin{bmatrix}11&8\\4&3\end{bmatrix}+a\begin{bmatrix}3&2\\1&1\end{bmatrix}+b\begin{bmatrix}1&0\\0&1\end{bmatrix}=0$
$\Rightarrow\ \begin{bmatrix}11&8\\4&3\end{bmatrix}+\begin{bmatrix}3a&2a\\a&a\end{bmatrix}+\begin{bmatrix}b&0\\0&b\end{bmatrix}=0$
$\Rightarrow\ \begin{bmatrix}11+3a+b&8+2a+0\\4+a+0&3+a+b\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\therefore$ We have $11 + 3a + b = 0 .....(1)$
$8 + 2a + 0 = 0$
$\Rightarrow 2a = -8$
$\Rightarrow a = -4$
Here $a = -4$ satisfies $4 + a + 0 = 0$ also, therefore $a = -4$
Putting $a = -4$ in eq. $(1), 11 - 12 + b = 0$
$\Rightarrow b - 1 = 0$
$\Rightarrow b = 1$
Here also $b = 1$ satisfies $3 + a + b = 0,$ therefore $b = 1$
Therefore$, a = -4$ and $b = 1$
View full question & answer→Question 1425 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 1435 Marks
$\text{If A} = \begin{bmatrix}3&1\\-1&2\end{bmatrix},$ show that $A^2 - 5A + 7I = 0$. Hence find $A^{-1}.$
AnswerGiven : $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$ $\therefore\ \text{A}^2=\text{A.A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$
$\text{L.H.S.}=\text{A}^2-5\text{A}+7\text{I}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}=\begin{bmatrix}8-15&5-5\\-5+5&3-10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}=\begin{bmatrix}-7&0\\0&-7\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$=\begin{bmatrix}-7+7&0+0\\0+0&-7+7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0=\text{R.H.S.}$
$\Rightarrow\ \text{A}^2-5\text{A}+7\text{I}=0$
To find: $A^{-1},$ multiplying eq. $(1)$ by $A^{-1}.$
$\Rightarrow\ \text{A}^2\text{A}^{-1}-5\text{A.A}^{-1}+7\text{I}_2\text{A}^{-1}=0.\text{A}^{-1}$
$\Rightarrow\ \text{A}-5\text{I}_2+7\text{A}^{-1}=0$
$\Rightarrow\ \ 7\text{A}^{-1}=-\text{A}+5\text{I}_2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}+5\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}+\begin{bmatrix}5&0\\0&5\end{bmatrix}=\begin{bmatrix}2&-1\\1&3\end{bmatrix}$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}$
View full question & answer→Question 1445 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 1455 Marks
For the matrix $\text{A}=\begin{bmatrix}1 & -1 & 1 \\2 & 3 & 0 \\ 18 & 2 & 10 \end{bmatrix},$ show that A (adjoint A) = 0.
Answer$\text{A}=\begin{bmatrix}1 & -1 & 1 \\2 & 3 & 0 \\ 18 & 2 & 10 \end{bmatrix}$
Now, $\text{C}_{11}=\begin{vmatrix}3 & 0 \\2 & 10 \end{vmatrix}= 30\ \text{C}_{12}=-\begin{vmatrix}2 & 0 \\18 & 10 \end{vmatrix}=-20$
$\text{C}_{13}=\begin{vmatrix}2 & 3 \\18 & 2 \end{vmatrix}=-50$
$\text{C}_{21}=\begin{vmatrix}-1 & 1 \\2 & 10 \end{vmatrix}=12\ \text{C}_{22}=-\begin{vmatrix}1 & 1 \\18 & 10 \end{vmatrix}=-8$
$ \text{C}_{23}=\begin{vmatrix}1 & -1 \\2 & 3 \end{vmatrix}=5$
$\text{C}_{31}=\begin{vmatrix}-1 & 1 \\3 & 0 \end{vmatrix}=-3\ \text{C}_{32}=-\begin{vmatrix}1 & 1 \\2 & 0 \end{vmatrix}=2$
$\text{C}_{33}=\begin{vmatrix}1 & -1 \\2 & 3 \end{vmatrix}=5$
$\text{adj A}=\begin{bmatrix}30 & -20 & -50\\12 & -8 & -20 \\ -3 & 2 & 5 \end{bmatrix}^\text{T}$
$=\begin{bmatrix}30 & 12 & -3\\-20 & -8 & 2 \\ -50 & -20 & 5 \end{bmatrix}$
$\therefore\ \text{A(adj A)}=\begin{bmatrix}-1 & -1 & 1\\2 & -8 & -2 \\ 18 & 2 & 10 \end{bmatrix}$
$=\begin{bmatrix}30 & 12 & -3\\-20 & -8 & 2 \\ -50 & -20 & 5 \end{bmatrix}$
$=\begin{bmatrix}30+20-50 & 12+18-20 & -3-2+5 \\60-60-0 & 24-24-0&-6+6+0 \\ 540-40-500 & 216-16-200 & -54+4+50 \end{bmatrix}$
$=\begin{bmatrix}0 & 0 & 0 \\0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
View full question & answer→Question 1465 Marks
Solve the following systems of homogeneous linear equations by matrix method:
3x + y - 2z = 0
x + y + z = 0
x - 2y + z = 0
AnswerThe given system of homogeneous equations can be written in matrix form as follows:
$\begin{bmatrix}3&1&-2\\1&1&1\\1&-2&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
or, $\text{AX}=\text{O}$
where, $\text{A}=\begin{bmatrix}3&1&-2\\1&1&1\\1&-2&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
Now,
$|\text{A}|=\begin{vmatrix}3&1&-2\\1&1&1\\1&-2&1\end{vmatrix}$
$=3(1+2)-1(1-1)-2(-2-1)$
$=9-0+6$
$=15\neq0$
So, the given system has only trivial solution, which is given below:
x = y = z = 0
View full question & answer→Question 1475 Marks
Solve the equation $\begin{vmatrix}x+a&x&x\\x&x+a&x\\x&x&x+a\end{vmatrix}=0,a\neq0$
Answer$\begin{vmatrix}x+a&x&x\\x&x+a&x\\x&x&x+a\end{vmatrix}=0$
Applying $R_{1 }\rightarrow R_1 + R_2 + R_3,$ we get:
$\begin{vmatrix}3x+a&3x+a&3x+a\\x&x+a&x\\x&x&x+a\end{vmatrix}=0$
$\Rightarrow\ \left(3x+a\right)\begin{vmatrix}1&1&1\\x&x+a&x\\x&x&x+a\end{vmatrix}=0$
Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$, we have:
$\left(3x+a\right)\begin{vmatrix}1&0&0\\x&a&0\\x&0&a\end{vmatrix}=0$
Expanding along $R_1,$ we have:
$(3x + a)[1 \times a^2] = 0$
$\Rightarrow a^{2 }(3x + a) = 0$
But $a \neq 0$.
Therefore, we have:
$3x + a = 0$
$\Rightarrow x =-\frac{a}{3}$
View full question & answer→Question 1485 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 1495 Marks
$\text{Let A} = \begin{bmatrix}3&7\\2&5\end{bmatrix}\text{and B} = \begin{bmatrix}6&8\\7&9\end{bmatrix}.$Verify that $(AB)^{-1} = B^{-1}A^{-1}.$
Answer$\text{Given}:\ \text{Matrix A}=\begin{bmatrix}3&7\\2&5\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}3&7\\2&5\end{vmatrix}=15-14=1\neq0$
$\therefore\ \text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj. A}=\frac{1}{1}\begin{bmatrix}5&-7\\-2&3\end{bmatrix}=\begin{bmatrix}5&-7\\-2&3\end{bmatrix}$
$\text{Matrix B}=\begin{bmatrix}6&8\\7&9\end{bmatrix}$ $\therefore\ \text{|B|}=\begin{vmatrix}6&8\\7&9\end{vmatrix}=54-56=-2\neq0$
$\therefore\ \text{B}^{-1}=\frac{1}{\text{|B|}}\text{adj. B}=\frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}$
$\text{Now}\ \text{AB}=\begin{bmatrix}3&7\\2&5\end{bmatrix}\begin{bmatrix}6&8\\7&9\end{bmatrix}=\begin{bmatrix}18+49&24+63\\12+35&16+45\end{bmatrix}=\begin{bmatrix}67&87\\47&61\end{bmatrix}$
$\therefore\ \text{|AB|}=\begin{vmatrix}67&87\\47&61\end{vmatrix}=67(61)-87(47)=4087-4089=-2\neq0$
$\text{Now}\ \text{L.H.S.}=\left(\text{AB}\right)^{-1}=\frac{1}{\text{|AB|}}\text{adj. (AB)}=\frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix} \dots\dots(1)$
$\text{R.H.S.}=\text{B}^{-1}\text{A}^{-1}=\frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}\begin{bmatrix}5&-7\\-2&3\end{bmatrix}=\frac{1}{-2}\begin{bmatrix}45+16&-63-24\\-35-12&49+18\end{bmatrix}$
$=\frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix} \dots\dots(2)$
From eq. $(1)$ and $(2),$ we get $\text{L.H.S. = R.H.S.}$
$\Rightarrow (AB)^{-1} = B^{-1}A^{-1}$
View full question & answer→Question 1505 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 $\text{z}=7$
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