Question 12 Marks
If A is a square matrix of order $3$ such that $|adj\ A| = 64$, find $|A|$.
AnswerFor any square matrix of order $n, |adj A| = |A|^{n-1}$
$\Rightarrow 64 = |A|^2 [\because |adj\ A| = 64] \Rightarrow|\text{A}|=\pm8$
View full question & answer→Question 22 Marks
If $\text{A}=\begin{bmatrix} 3 & 1 \\ 2 & -3 \end{bmatrix},$ then find |adj A|.
Answer$\text{A}=\begin{bmatrix} 3 & 1 \\ 2 & -3 \end{bmatrix}$Now, $|adj\ A| = |A|^{n-1}$ .....(i)
Now, $|A| = -9 - 2 = -11$
So, $|adj\ A| = (-11)^{2-1}$
$= (-11)^1$
$= -11$
Hence, $|adj\ A| = -11$
View full question & answer→Question 32 Marks
Write the adjoint of the matrix $\text{A}=\begin{bmatrix} -3 & 4 \\ 7 & -2 \end{bmatrix}$.
AnswerLet $C_{ij}$ be a cofactor of $a_{ij}$ in A. Now, $C_{11} = -2 C_{12} = -7 C_{21} = -4 C_{22} = -3$
$\therefore\ \text{adj A}=\begin{bmatrix} -2 & -7 \\ -4 & -3 \end{bmatrix}^\text{T}=\begin{bmatrix} -2 & -4 \\ -7 & -3 \end{bmatrix}$
View full question & answer→Question 42 Marks
If $A$ is a non-singular symmetric matrix, write whether $A^{-1}$ is symmetric or skew-symmetric.
AnswerLet $A$ be an invertible symmetric matrix.
Then,$|\text{A}|\neq0\text{ and }\text{A}^\text{T}=\text{A}$
Now, $(A^{-1})^T = (A^T)^{-1}$
$\Rightarrow (A^{-1})^T = A^{-1} [ \because A^T = A]$
Thus, $A^{-1}$ is symmetric matrix.
View full question & answer→Question 52 Marks
If A is a square matrix of order $3$ such that $|A| = 5$, write the value of $|adj A|$.
AnswerFor any square matrix of order n,
$|adj\ A| = |A|^{n-1}$
$\Rightarrow |adj\ A| = |A|^2= 5^2= 25$
View full question & answer→Question 62 Marks
If $\text{A}=\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix},\text{B}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ find adj (AB).
Answer$\text{A}\times\text{B}=\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}$A × B is a non-singular matrix. Therefore, it is invertible.
Let $C_{ij}$ be a cofactor of $a_{ij}$ in A.
The cofactors of element A are given by
$C_{11} = d$
$C_{12} = -c$
$C_{21} = -b$
$C_{22} = a$
$\therefore\ \text{adj A}=\begin{bmatrix} \text{d} & -\text{c} \\ -\text{b} & \text{a} \end{bmatrix}^\text{T}=\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
View full question & answer→Question 72 Marks
If $A$ is a square matrix of order $3$ such that $|A| = 2,$ then write the value of adj $(adj\ A).$
AnswerFor any square matrix $A$, we have
adj $(adj\ A) = |A|^{n-2} A$
$\Rightarrow$ adj $(adj\ A) = 2A [\because |A| = 2$ and $n = 3]$
View full question & answer→Question 82 Marks
If A is a square matrix, then write the matrix adj $(A^T) − (adj\ A)^T$.
AnswerIn a non-singular matrix, adj $A^{\top}=(\operatorname{adj} A)^{\top}$.
$\Rightarrow\left(\operatorname{adj} A^{\top}\right)-(\operatorname{adj} A)^{\top}=$ Null matrix
View full question & answer→Question 92 Marks
If $\text{A}=\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix},$ write $A^{-1}$ in terms of $A$.
Answer$|\text{A}|=\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}=-19\neq0$A is a non-singular matrix. Therefore, it is invertible.
Let $C_{ij}$ be a cofactor of $a_{ij}$ in A.
The cofactors of element A are given by
$C_{11}= -2$
$C_{12} = -5$
$C_{21} = -3$
$C_{22} = 2$
$\text{adj A}=\begin{bmatrix} -2 & -5 \\ -3 & 2 \end{bmatrix}^\text{T}=\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix},$
$\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\begin{bmatrix} \frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & \frac{-2}{19} \end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{19}\text{A}$
View full question & answer→Question 102 Marks
Find the inverse of the matrix $\begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$.
Answer$|\text{A}|=\begin{vmatrix} 3 & -2 \\ -7 & 5 \end{vmatrix}=1\neq0$A is a non-singular matrix. Therefore, is is invertible.
Let $C_{ij}$ be a cofactor of $a_{ij}$ in A.
The cofactors of element A are given by
$C_{11}= 5$
$C_{12} = 7$
$C_{21}= 2$
$C_{22} = 3$
$\therefore\ \text{A}^{-1}=\frac{1}{|\text{A}|}\begin{bmatrix} 5 & 7 \\ 2 & 3 \end{bmatrix}^\text{T}=\begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}$
View full question & answer→Question 112 Marks
If $A$ is a square matrix of order $3$ such that adj $(2A) = k$ adj $(A),$ then write the value of $k.$
AnswerFor any matrix A of order n, adj $(\lambda\text{A})=\lambda^{\text{n}-1}=\lambda^{\text{n}-1}$ (adj A) where $\lambda$ is a constant.
Thus, for matrix A of order 3, we have
$adj (2A) = 2^{3-1} (adj A)$
$\Rightarrow\ adj\ (2A) = 2^2 (adj\ A)$
$\Rightarrow\ adj\ (2A) = 4(adj) (A)$
$\Rightarrow\ k\ adj (A) = 4\ adj (A) [ \because\ adj (2A) = k\ adj\ (A)]$
$\Rightarrow k = 4$
View full question & answer→Question 122 Marks
Find the inverse of the matrix $\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$.
Answer$\text{A}=\begin{bmatrix} 1 & -3 \\ 2 & 0 \end{bmatrix}$Cofactors of A are:
$C_{11} = 0, C_{21} = 3$
$C_{12} = -2 C_{22} = 1$
$\text{Adj A}=\begin{bmatrix} 0 & 3 \\ -2 & 1 \end{bmatrix}$
View full question & answer→Question 132 Marks
If A is a non-singular square matrix such that $|A| = 10$, find $|A^{-1}|$.
Answer$\big|\text{A}^{-1}\big|=\Big|\frac{1}{\text{A}}\Big|$$=\Big|\frac{1}{\text{A}}\Big|$
$=\frac{1}{10}\ \big[\because|\text{A}|=10\text{ (Given})\big]$
Hence, $\big|\text{A}^{-1}\big|=\frac{1}{10}$
View full question & answer→Question 142 Marks
Find the adjoint of the following matrices:$\text{C}=\begin{bmatrix} \cos\alpha & \sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$
Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.
Answer$\text{adjoint C}=\begin{bmatrix} \cos\alpha & -\sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$$\text{(adjoint C)C}=\begin{bmatrix}\cos^2\alpha-\sin^2\alpha & 0 \\ 0 & \cos^2\alpha-\sin^2\alpha \end{bmatrix}$
$|\text{C}|=\cos^2\alpha-\sin^2\alpha$
$|\text{C}|\text{I}=\begin{bmatrix}\cos^2\alpha-\sin^2\alpha & 0 \\ 0 & \cos^2\alpha-\sin^2\alpha \end{bmatrix}$
$\text{C(adjoint C)}=\begin{bmatrix}\cos^2\alpha-\sin^2\alpha & 0 \\ 0 & \cos^2\alpha-\sin^2\alpha \end{bmatrix}$
$\therefore\ \text{(adjoint C)}=|\text{C}|\text{I}=\text{C(adjoint C)}$
Hence verified.
View full question & answer→Question 152 Marks
If A is a square matrix such that A (adj A) = 5I, where I denotes the identity matrix of the same order. Then, find the value of |A|.
AnswerA (adj A) = 5I (Given) |A| I = 5I [$\because$ A(adj A) = |A|I]$\therefore |\text{A}|=5$
View full question & answer→Question 162 Marks
If A is a non-singular square matrix such that $\text{A}^{-1}=\begin{bmatrix} 5 & 3 \\ -2 & -1 \end{bmatrix},$ then find $(A^T)^{-1}$.
AnswerFor any invertible matrix A. $(A^T)^{-1} = (A^{-1})^T$ We have $\text{A}^{-1}=\begin{bmatrix} 5 & 3 \\ -2 & -1 \end{bmatrix}$$\Rightarrow(\text{A}^\text{T})^{-1}=\begin{bmatrix} 5 & -2 \\ 3 & -1 \end{bmatrix}$
View full question & answer→Question 172 Marks
If A is a square matrix such that $\text{A (adj A)}=\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix},$ then write the value of |adj A|.
AnswerGiven,$\text{A (adj A)}=\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
$\Rightarrow|\text{A}|\text{I}_\text{n}=5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow|\text{A}|=5$
Now, $|adj\ A| = |A|^{n-1} = 5^{3-1} = 25$.
View full question & answer→Question 182 Marks
Write $A^{-1}$ for $\text{A}=\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$
Answer$|\text{A}|=\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}=1\neq0$Let $C_{ij}$ be the cofactor of $a_{ij}$ in A.
The cofactors of element A are given by
$C_{11} = 3$
$C_{12} = -1$
$C_{21} = -5$
$C_{22} = 2$
$\text{adj A}=\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}^\text{T}=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$
$|\text{A}|=6-5=1$
$\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$
View full question & answer→Question 192 Marks
If A is symmetric matrix, write whether $A^T$ is symmetric or skew-symmetric.
AnswerFor any symmetric matrix, $A^T = A$.
Hence, $A^T$ is also symmetric.
View full question & answer→Question 202 Marks
If A is an invertible matrix such that $|A^{-1}| = 2$, find the value of $|A|$.
Answer$\big|\text{A}^{-1}\big|=2$$\Big|\frac{1}{\text{A}}\Big|=2$
$\frac{1}{|\text{A}|}=2$
$\therefore|\text{A}|=\frac{1}{2}$
Hence, $|\text{A}|=\frac{1}{2}$
View full question & answer→Question 212 Marks
Use elementary column operation $C_2 → C_2 + 2C_1$ in the following matrix equation:$\begin{pmatrix} 2 & 1 \\ 2 & 0 \end{pmatrix}=\begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$
Answer$\begin{pmatrix} 2 & 1 \\ 2 & 0 \end{pmatrix}=\begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$Applying $C_2 → C_2 + 2C_1$
$\begin{pmatrix} 2 & 5 \\ 2 & 4 \end{pmatrix}=\begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix}$
View full question & answer→Question 222 Marks
In the following matrix equation use elementary operation $R_2 → R_2 + R_1$ and the equation thus obtained:$\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix}\begin{bmatrix} 8 & -3 \\ 9 & -4 \end{bmatrix}$
Answer$\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix}\begin{bmatrix} 8 & -3 \\ 9 & -4 \end{bmatrix}$By applying elementary operation $R_2 → R_2 + R_1$, we get
$\begin{bmatrix} 2 & 3 \\ 3 & 7 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix}\begin{bmatrix} 8 & -3 \\ 17 & -7 \end{bmatrix}$
(Every row operation is equlvalent to left-multiplication be an elementary matrix.)
View full question & answer→Question 232 Marks
Let A be a $3 \times 3$ square matrix, such that $A (adj\ A) = 2I,$ where I is the identity matrix. Write the value of $|adj\ A|.$
Answer$\because A(adj\ (A)) = |A|I2I = |A|I ($Given $A(adj\ A) = 2I)$
$|A| = 2$
Also, $|adj\ A| = |A|^{n-1}$
$= (2)^{3-1}$
$= (2)^2$
$= 4$
$|adj\ A| = 4$
View full question & answer→Question 242 Marks
If $\text{A}=\begin{bmatrix} 1 & -3 \\ 2 & 0 \end{bmatrix},$ write adj A.
Answer$|\text{A}|=\begin{bmatrix} 1 & -3 \\ 2 & 0 \end{bmatrix}=6\neq0$A is a non-singular matrix. Therefore, it is invertible.
Let $C_{ij}$ be a cofactor of $a_{ij}$ in A.
The cofactors of element A are given by
$C_{11} = 0$
$C_{12} = -2$
$C_{21} = 3$
$C_{22} = 1$
$\therefore\ \text{adj A}=\text{A}=\begin{bmatrix} 0 & -2 \\ 3 & 1 \end{bmatrix}^\text{T}=\begin{bmatrix} 0 & 3 \\ -2 & 1 \end{bmatrix}$
View full question & answer→Question 252 Marks
Find the adjoint of the following matrices:$\begin{bmatrix}1 & \frac{\tan\alpha}{2} \\ -\frac{\tan\alpha}{2} & 1 \end{bmatrix}$
Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.
Answer$\text{D}=\begin{bmatrix}1 & \frac{\tan\alpha}{2} \\ -\frac{\tan\alpha}{2} & 1 \end{bmatrix}$$\text{adjoint D}=\begin{bmatrix}1 & -\frac{\tan\alpha}{2} \\ \frac{\tan\alpha}{2} & 1 \end{bmatrix}$
$(\text{adjoint D)D}=\begin{bmatrix} 1+\tan^2\frac{\alpha}{2} & 0 \\ 0 &1+\tan^2\frac{\alpha}{2} \end{bmatrix}$
$|\text{D}|=1+\tan^2\frac{\alpha}{2}$
$|\text{D}|\text{I}=\begin{bmatrix} 1+\tan^2\frac{\alpha}{2} & 0 \\ 0 &1+\tan^2\frac{\alpha}{2} \end{bmatrix}$
View full question & answer→Question 262 Marks
If $C_{ij}$ is the cofactor of the element $a_{ij}$ of the matrix $\text{A}=\begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix},$ then write the value of $a_{32}C_{32}$.
AnswerIn the given matrix $\text{A}=\begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix}$
$C_{32} = (-1)^{3+2} (8 - 30) = 22$
Therefore, $a_{32}C_{32} = 5 \times 22 = 110$.
Hence, the value of $a_{32}C_{32}$ is $110$.
View full question & answer→Question 272 Marks
$A, B, C$ are three non-null square matrices of the same order, write the condition on $A$ such that $AB = AC \Rightarrow B = C.$
AnswerConsider $AB = AC$. On multiplying both sides by $A^{-1},$
we get $AA^{-1}B = AA^{-1}$
$\Rightarrow IB = IC [$Because $AA^{-1} = I$ where I is the identity matrix$] \Rightarrow B = C$
Therefore, the required condition is $A$ must be invertible or $|\text{A}|\neq0$.
View full question & answer→Question 282 Marks
Let $A$ be a square matrix such that $A^2 - A + I = 0,$ then write $A^{-1}$ interms of $A.$
Answer$A^2 - A + I = 0$
$Px-$multiplying with $A^{-1},$
$(A^{-1} A) - (A^{-1} A) + A^{-1}I = 0$
$IA - I + A^{-1} = 0$
$A^{-1} = I - A$
Hence, $A^{-1} = I - A$
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