Adjoint and inverse of matrices Questions in English

(B) Let $A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$. We are given $|A| = 5$.
The given determinant is the determinant of the cofactor matrix of $A$,which is denoted as $|adj(A)|$.
We know the property $|adj(A)| = |A|^{n-1}$,where $n$ is the order of the matrix.
Here,$n = 3$,so $|adj(A)| = |A|^{3-1} = |A|^2$.
Substituting the given value,$|adj(A)| = 5^2 = 25$.

(B) Given that $AB = O$ and $B$ is a non-singular matrix.
Since $B$ is non-singular,its determinant $|B| \neq 0$,which implies that the inverse matrix $B^{-1}$ exists.
Now,multiply both sides of the equation $AB = O$ by $B^{-1}$ on the right side:
$(AB)B^{-1} = OB^{-1}$
Using the associative property of matrix multiplication:
$A(BB^{-1}) = O$
Since $BB^{-1} = I$,where $I$ is the identity matrix:
$AI = O$
Therefore,$A = O$.

(D) matrix $A$ is invertible if and only if its determinant $|A| \neq 0$.
Calculating the determinant of matrix $A$ by expanding along the second column:
$|A| = 0 \cdot \begin{vmatrix} 2 & 3 \\ k & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -k \\ k & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} 1 & -k \\ 2 & 3 \end{vmatrix}$
$|A| = 1 \cdot (1 - (-k^2)) = 1 + k^2$.
Since $k^2 \geq 0$ for all real $k$,$1 + k^2 \geq 1$.
Therefore,$|A|$ is always greater than or equal to $1$,which means $|A|$ can never be $0$.
Hence,the matrix $A$ is invertible for all real values of $k$.

(C) Let $A = \left[ {\begin{array}{*{20}{c}}3&{ - 2}&{ - 1}\\{ - 4}&1&{ - 1}\\2&0&1\end{array}} \right]$.
First,we calculate the determinant $|A|$:
$|A| = 3(1(1) - (-1)(0)) - (-2)((-4)(1) - (-1)(2)) + (-1)((-4)(0) - 1(2))$
$|A| = 3(1) + 2(-4 + 2) - 1(0 - 2) = 3 - 4 + 2 = 1$.
Next,we find the matrix of cofactors $C_{ij}$:
$C_{11} = +((1)(1) - (-1)(0)) = 1$
$C_{12} = -((-4)(1) - (-1)(2)) = -(-4 + 2) = 2$
$C_{13} = +((-4)(0) - (1)(2)) = -2$
$C_{21} = -((-2)(1) - (-1)(0)) = -(-2) = 2$
$C_{22} = +((3)(1) - (-1)(2)) = 3 + 2 = 5$
$C_{23} = -((3)(0) - (-2)(2)) = -(4) = -4$
$C_{31} = +((-2)(-1) - (1)(-1)) = 2 + 1 = 3$
$C_{32} = -((3)(-1) - (-4)(-1)) = -(-3 - 4) = 7$
$C_{33} = +((3)(1) - (-2)(-4)) = 3 - 8 = -5$
The cofactor matrix is $\left[ {\begin{array}{*{20}{c}}1&2&{-2}\\2&5&{-4}\\3&7&{-5}\end{array}} \right]$.
The adjoint matrix $adj(A)$ is the transpose of the cofactor matrix:
$adj(A) = \left[ {\begin{array}{*{20}{c}}1&2&3\\2&5&7\\{-2}&{-4}&{-5}\end{array}} \right]$.
Since $A^{-1} = \frac{1}{|A|} adj(A)$ and $|A| = 1$,we have:
$A^{-1} = \left[ {\begin{array}{*{20}{c}}1&2&3\\2&5&7\\{-2}&{-4}&{-5}\end{array}} \right]$.
Thus,the correct option is $C$.

(D) For any two non-singular matrices $A$ and $B$ of the same order,the inverse of their product is given by the reversal law of inverses.
Specifically,$(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = A(I)A^{-1} = AA^{-1} = I$.
Similarly,$(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}(I)B = B^{-1}B = I$.
Therefore,$(AB)^{-1} = B^{-1}A^{-1}$.

(A) To find the adjoint of matrix $N$,we calculate the cofactor matrix $C$ and then take its transpose $C^T$.
The matrix is $N = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$.
The cofactors are calculated as follows:
$c_{11} = +((0)(3) - (1)(4)) = -4$
$c_{12} = -((1)(3) - (1)(4)) = -(-1) = 1$
$c_{13} = +((1)(4) - (0)(4)) = 4$
$c_{21} = -((-3)(3) - (-3)(4)) = -(-9 + 12) = -3$
$c_{22} = +((-4)(3) - (-3)(4)) = -12 + 12 = 0$
$c_{23} = -((-4)(4) - (-3)(4)) = -(-16 + 12) = 4$
$c_{31} = +((-3)(1) - (-3)(0)) = -3$
$c_{32} = -((-4)(1) - (-3)(1)) = -(-4 + 3) = 1$
$c_{33} = +((-4)(0) - (-3)(1)) = 3$
The cofactor matrix is $C = \begin{bmatrix} -4 & 1 & 4 \\ -3 & 0 & 4 \\ -3 & 1 & 3 \end{bmatrix}$.
The adjoint is the transpose of the cofactor matrix:
$adj(N) = C^T = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix} = N$.

(B) Let $I$ be the identity matrix of order $3 \times 3$,given by $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Then,$kI = \begin{bmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{bmatrix}$.
The adjoint of a diagonal matrix $D = \text{diag}(a, b, c)$ is given by $\text{diag}(bc, ac, ab)$.
Here,$adj(kI) = \begin{bmatrix} k \times k & 0 & 0 \\ 0 & k \times k & 0 \\ 0 & 0 & k \times k \end{bmatrix} = \begin{bmatrix} k^2 & 0 & 0 \\ 0 & k^2 & 0 \\ 0 & 0 & k^2 \end{bmatrix} = k^2 I$.
Alternatively,using the property $adj(kA) = k^{n-1} adj(A)$ where $n$ is the order of the matrix,for $A=I$ and $n=3$,we get $adj(kI) = k^{3-1} adj(I) = k^2 I$.

(B) The adjoint of a matrix $A$ is denoted by $adj(A)$.
For any square matrix $A$ of order $n \times n$,the property of the adjoint of an adjoint matrix is given by the formula:
$adj(adj \,A) = |A|^{n - 2}A$
Thus,the correct option is $B$.

(B) Given $A = \begin{bmatrix} i & 0 \\ 0 & i/2 \end{bmatrix}$.
The determinant of $A$ is $|A| = (i)(i/2) - (0)(0) = i^2/2 = -1/2$.
The adjoint of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Thus,$adj(A) = \begin{bmatrix} i/2 & 0 \\ 0 & i \end{bmatrix}$.
The inverse is given by $A^{-1} = \frac{1}{|A|} adj(A)$.
$A^{-1} = \frac{1}{-1/2} \begin{bmatrix} i/2 & 0 \\ 0 & i \end{bmatrix} = -2 \begin{bmatrix} i/2 & 0 \\ 0 & i \end{bmatrix} = \begin{bmatrix} -i & 0 \\ 0 & -2i \end{bmatrix}$.

(C) We know that for any square matrix $A$ of order $n$,the fundamental property of the adjoint matrix is given by:
$A(\text{adj } A) = (\text{adj } A)A = |A|I$,where $|A|$ is the determinant of $A$ and $I$ is the identity matrix of the same order as $A$.
Since $A$ is a non-singular matrix,$|A| \neq 0$.
Thus,the correct expression is $|A|I$.

(B) Let $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$.
First,we calculate the determinant $|A| = 1(1-0) - 2(2-0) + 1(0 - (-1)) = 1 - 4 + 1 = -2$.
The inverse is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
The element in the $2^{nd}$ row and $3^{rd}$ column of $A^{-1}$ is $\frac{1}{|A|} \times C_{32}$,where $C_{32}$ is the cofactor of the element in the $3^{rd}$ row and $2^{nd}$ column of $A$.
The cofactor $C_{32} = (-1)^{3+2} M_{32} = -1 \times \begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix} = -1 \times (0 - 2) = 2$.
Thus,the required element is $\frac{C_{32}}{|A|} = \frac{2}{-2} = -1$.

(A) The inverse of a matrix $A$ is given by the formula $A^{-1} = \frac{adj(A)}{|A|}$.
First,we calculate the determinant of $A$:
$|A| = \begin{vmatrix} a & c \\ d & b \end{vmatrix} = ab - cd$.
Next,we find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$adj(A) = \begin{bmatrix} b & -c \\ -d & a \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{ab - cd} \begin{bmatrix} b & -c \\ -d & a \end{bmatrix}$.

(B) The given matrix is the identity matrix of order $3$,denoted by $I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
By the definition of the inverse of a matrix,for any square matrix $A$,if $AB = BA = I$,then $B$ is the inverse of $A$,denoted as $A^{-1}$.
For the identity matrix $I$,we know that $I \times I = I$.
Therefore,the inverse of the identity matrix is the identity matrix itself,i.e.,$I^{-1} = I$.
Thus,the inverse of $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

(A) Let $A = \begin{bmatrix} 2 & -3 \\ -4 & 2 \end{bmatrix}$.
First,we find the determinant $|A| = (2)(2) - (-3)(-4) = 4 - 12 = -8$.
Since $|A| \neq 0$,the inverse exists.
The adjoint of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Thus,$adj(A) = \begin{bmatrix} 2 & 3 \\ 4 & 2 \end{bmatrix}$.
The inverse is given by $A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-8} \begin{bmatrix} 2 & 3 \\ 4 & 2 \end{bmatrix} = -\frac{1}{8} \begin{bmatrix} 2 & 3 \\ 4 & 2 \end{bmatrix}$.

(D) To find the adjoint of matrix $A$,we first find the matrix of cofactors $C_{ij}$.
Given $A = \begin{bmatrix} 1 & 0 & 0 \\ 5 & 2 & 0 \\ -1 & 6 & 1 \end{bmatrix}$.
The cofactors are:
$C_{11} = +\begin{vmatrix} 2 & 0 \\ 6 & 1 \end{vmatrix} = 2 - 0 = 2$
$C_{12} = -\begin{vmatrix} 5 & 0 \\ -1 & 1 \end{vmatrix} = -(5 - 0) = -5$
$C_{13} = +\begin{vmatrix} 5 & 2 \\ -1 & 6 \end{vmatrix} = 30 - (-2) = 32$
$C_{21} = -\begin{vmatrix} 0 & 0 \\ 6 & 1 \end{vmatrix} = -(0 - 0) = 0$
$C_{22} = +\begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix} = 1 - 0 = 1$
$C_{23} = -\begin{vmatrix} 1 & 0 \\ -1 & 6 \end{vmatrix} = -(6 - 0) = -6$
$C_{31} = +\begin{vmatrix} 0 & 0 \\ 2 & 0 \end{vmatrix} = 0 - 0 = 0$
$C_{32} = -\begin{vmatrix} 1 & 0 \\ 5 & 0 \end{vmatrix} = -(0 - 0) = 0$
$C_{33} = +\begin{vmatrix} 1 & 0 \\ 5 & 2 \end{vmatrix} = 2 - 0 = 2$
Thus,the cofactor matrix is $C = \begin{bmatrix} 2 & -5 & 32 \\ 0 & 1 & -6 \\ 0 & 0 & 2 \end{bmatrix}$.
The adjoint of $A$ is the transpose of the cofactor matrix:
$adj(A) = C^T = \begin{bmatrix} 2 & 0 & 0 \\ -5 & 1 & 0 \\ 32 & -6 & 2 \end{bmatrix}$.
Comparing this with the given options,none of them match. Therefore,the correct option is $(D)$.

(A) We know that for any square matrix $A$ of order $n$,$A(\text{adj } A) = |A|I$,where $I$ is the identity matrix of order $n$.
Given $A = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$.
First,calculate the determinant $|A| = (3 \times 4) - (2 \times 1) = 12 - 2 = 10$.
Since $A$ is a $2 \times 2$ matrix,$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Therefore,$A(\text{adj } A) = |A|I = 10 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$.

(B) Given $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$.
We know that for any square matrix $A$,$A \cdot \text{adj}(A) = |A| I$,where $I$ is the identity matrix.
First,calculate the determinant of $A$:
$|A| = (\cos \alpha)(\cos \alpha) - (\sin \alpha)(-\sin \alpha) = \cos^2 \alpha + \sin^2 \alpha = 1$.
Thus,$A \cdot \text{adj}(A) = 1 \cdot \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Comparing this with the given matrix $\begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$,we get $k = 1$.

(A) Given the equation: $3A^3 + 2A^2 + 5A + I = 0$
Subtract $I$ from both sides: $3A^3 + 2A^2 + 5A = -I$
Multiply both sides by $A^{-1}$ from the right: $(3A^3 + 2A^2 + 5A)A^{-1} = -I A^{-1}$
Using the property $AA^{-1} = I$: $3A^2(AA^{-1}) + 2A(AA^{-1}) + 5(AA^{-1}) = -A^{-1}$
$3A^2(I) + 2A(I) + 5(I) = -A^{-1}$
$3A^2 + 2A + 5I = -A^{-1}$
Therefore,$A^{-1} = -(3A^2 + 2A + 5I)$

(D) Let us analyze each statement:
$(i)$ If $A$ is a symmetric matrix,then $A^T = A$. The adjoint of $A$ is defined as the transpose of the cofactor matrix. It can be shown that $adj(A^T) = (adj\,A)^T$. Since $A^T = A$,we have $adj(A) = (adj\,A)^T$,which means $adj(A)$ is symmetric. This statement is true.
$(ii)$ If $A = I$ (unit matrix),then $adj(I) = |I|I^{-1} = 1 \times I = I$. Thus,the adjoint of a unit matrix is a unit matrix. This statement is true.
$(iii)$ By the fundamental property of the adjoint of a matrix,$A(adj\,A) = (adj\,A)A = |A|I$. This statement is true.
$(iv)$ If $A$ is a diagonal matrix,its cofactors will result in another diagonal matrix. Thus,the adjoint of a diagonal matrix is a diagonal matrix. This statement is true.
Since all statements $(i), (ii), (iii),$ and $(iv)$ are correct,none of the statements are incorrect. Therefore,the correct option is $(d)$.

(B) Let $A = \left[ {\begin{array}{*{20}{c}}1&3\\3&{10}\end{array}} \right]$.
To find the inverse $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
First,calculate the determinant $|A| = (1 \times 10) - (3 \times 3) = 10 - 9 = 1$.
Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements: $\text{adj}(A) = \left[ {\begin{array}{*{20}{c}}{10}&{-3}\\{-3}&1\end{array}} \right]$.
Thus,$A^{-1} = \frac{1}{1} \left[ {\begin{array}{*{20}{c}}{10}&{-3}\\{-3}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{10}&{-3}\\{-3}&1\end{array}} \right]$.
Therefore,the correct option is $B$.

(A) Let $A$ be a symmetric matrix.
By definition of a symmetric matrix,$A^T = A$.
We know that $A A^{-1} = I$.
Taking the transpose on both sides,we get $(A A^{-1})^T = I^T$.
Using the property $(AB)^T = B^T A^T$,we have $(A^{-1})^T A^T = I$.
Since $A^T = A$,this becomes $(A^{-1})^T A = I$.
Multiplying both sides by $A^{-1}$ on the right,we get $(A^{-1})^T A A^{-1} = I A^{-1}$.
$(A^{-1})^T I = A^{-1}$.
$(A^{-1})^T = A^{-1}$.
Since the transpose of $A^{-1}$ is equal to $A^{-1}$,the inverse of a symmetric matrix is symmetric.

(A) Let $A = \left[ {\begin{array}{*{20}{c}}{ - 6}&5\\{ - 7}&6\end{array}} \right]$.
To find $A^{-1}$,we use the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
First,calculate the determinant $|A| = (-6)(6) - (5)(-7) = -36 + 35 = -1$.
Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements: $\text{adj}(A) = \left[ {\begin{array}{*{20}{c}}6&{-5}\\7&{-6}\end{array}} \right]$.
Therefore,$A^{-1} = \frac{1}{-1} \left[ {\begin{array}{*{20}{c}}6&{-5}\\7&{-6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}-6&5\\-7&6\end{array}} \right]$.

(B) Let $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$. The adjoint of $A$ is the transpose of the cofactor matrix,$adj(A) = [C_{ij}]^T = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix}$.
Calculating the cofactors:
$C_{11} = +\begin{vmatrix} 2 & -3 \\ -1 & 3 \end{vmatrix} = (6 - 3) = 3$
$C_{12} = -\begin{vmatrix} 1 & -3 \\ 2 & 3 \end{vmatrix} = -(3 + 6) = -9$
$C_{13} = +\begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (-1 - 4) = -5$
$C_{21} = -\begin{vmatrix} 1 & 1 \\ -1 & 3 \end{vmatrix} = -(3 + 1) = -4$
$C_{22} = +\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = (3 - 2) = 1$
$C_{23} = -\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = -(-1 - 2) = 3$
$C_{31} = +\begin{vmatrix} 1 & 1 \\ 2 & -3 \end{vmatrix} = (-3 - 2) = -5$
$C_{32} = -\begin{vmatrix} 1 & 1 \\ 1 & -3 \end{vmatrix} = -(-3 - 1) = 4$
$C_{33} = +\begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = (2 - 1) = 1$
Thus,$adj(A) = \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix}$. The correct option is $B$.

(B) To find the inverse of a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,we use the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
Given $A = \begin{bmatrix} 5 & 2 \\ 3 & 1 \end{bmatrix}$.
First,calculate the determinant $|A| = (5)(1) - (2)(3) = 5 - 6 = -1$.
Next,find the adjoint matrix $\text{adj}(A)$ by swapping the diagonal elements and changing the signs of the off-diagonal elements: $\text{adj}(A) = \begin{bmatrix} 1 & -2 \\ -3 & 5 \end{bmatrix}$.
Now,$A^{-1} = \frac{1}{-1} \begin{bmatrix} 1 & -2 \\ -3 & 5 \end{bmatrix} = \begin{bmatrix} -1 & 2 \\ 3 & -5 \end{bmatrix}$.
Thus,the correct option is $B$.

(A) Let $A = \begin{bmatrix} 3 & -2 \\ 1 & 4 \end{bmatrix}$.
The determinant of $A$ is $|A| = (3)(4) - (-2)(1) = 12 + 2 = 14$.
The adjoint of $A$ is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements: $adj(A) = \begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix}$.
The inverse of the matrix is given by $A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{14} \begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} \frac{4}{14} & \frac{2}{14} \\ \frac{-1}{14} & \frac{3}{14} \end{bmatrix}$.

(A) Given $A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
First,we find the determinant $|A| = 0(0-0) - 1(1-0) + 0(0-0) = -1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
We find the cofactor matrix $C$ where $C_{ij} = (-1)^{i+j} M_{ij}$.
$C_{11} = +(0-0) = 0, C_{12} = -(1-0) = -1, C_{13} = +(0-0) = 0$.
$C_{21} = -(1-0) = -1, C_{22} = +(0-0) = 0, C_{23} = -(0-0) = 0$.
$C_{31} = +(0-0) = 0, C_{32} = -(0-0) = 0, C_{33} = +(0-1) = -1$.
Thus,$adj(A) = C^T = \begin{bmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix}$.
Finally,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-1} \begin{bmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} = A$.

(A) matrix $A$ is singular if its determinant is zero,i.e.,$|A| = 0$.
We know the property of the adjoint of a matrix: $|\text{adj } A| = |A|^{n-1}$,where $n$ is the order of the matrix $A$.
If $n > 1$,then since $|A| = 0$,we have $|\text{adj } A| = 0^{n-1} = 0$.
Therefore,$\text{adj } A$ is also a singular matrix.

(B) Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}$.
First,we calculate the determinant $|A| = 1(1 \times 1 - 2 \times 0) - 2(0 \times 1 - 2 \times 0) + 3(0 \times 0 - 1 \times 0) = 1(1) = 1$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
Next,we find the matrix of cofactors $C_{ij}$:
$C_{11} = +|\begin{smallmatrix} 1 & 2 \\ 0 & 1 \end{smallmatrix}| = 1, C_{12} = -|\begin{smallmatrix} 0 & 2 \\ 0 & 1 \end{smallmatrix}| = 0, C_{13} = +|\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}| = 0$
$C_{21} = -|\begin{smallmatrix} 2 & 3 \\ 0 & 1 \end{smallmatrix}| = -2, C_{22} = +|\begin{smallmatrix} 1 & 3 \\ 0 & 1 \end{smallmatrix}| = 1, C_{23} = -|\begin{smallmatrix} 1 & 2 \\ 0 & 0 \end{smallmatrix}| = 0$
$C_{31} = +|\begin{smallmatrix} 2 & 3 \\ 1 & 2 \end{smallmatrix}| = 4-3 = 1, C_{32} = -|\begin{smallmatrix} 1 & 3 \\ 0 & 2 \end{smallmatrix}| = -2, C_{33} = +|\begin{smallmatrix} 1 & 2 \\ 0 & 1 \end{smallmatrix}| = 1$
Thus,$Adj(A) = \begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix}$.
Finally,$A^{-1} = \frac{1}{|A|} Adj(A) = \frac{1}{1} \begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix}$.

(C) For option $A$: $A$ non-singular square matrix always has a unique inverse,so this statement is false.
For option $B$: By definition,a non-singular matrix has a non-zero determinant $(|A| \neq 0)$,so this statement is false.
For option $C$: If $A' = A$,then the number of rows must equal the number of columns,which implies $A$ is a square matrix. Thus,this statement is true.
For option $D$: We know that $A \cdot \text{adj } A = |A| I_n$. Taking the determinant on both sides,$|A \cdot \text{adj } A| = ||A| I_n| = |A|^n |I_n| = |A|^n$. The given expression $|A|^{n-1}$ is incorrect. Therefore,this statement is false.

(B) Given the matrix $A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 2 & 0 & 1 \end{bmatrix}$.
To find the adjoint of $A$,we calculate the cofactor matrix $C = [C_{ij}]$ and then take its transpose.
The cofactors are calculated as follows:
$C_{11} = +\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 - 0 = 1$
$C_{12} = -\begin{vmatrix} 0 & 2 \\ 2 & 1 \end{vmatrix} = -(0 - 4) = 4$
$C_{13} = +\begin{vmatrix} 0 & 1 \\ 2 & 0 \end{vmatrix} = 0 - 2 = -2$
$C_{21} = -\begin{vmatrix} 2 & 0 \\ 0 & 1 \end{vmatrix} = -(2 - 0) = -2$
$C_{22} = +\begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} = 1 - 0 = 1$
$C_{23} = -\begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} = -(0 - 4) = 4$
$C_{31} = +\begin{vmatrix} 2 & 0 \\ 1 & 2 \end{vmatrix} = 4 - 0 = 4$
$C_{32} = -\begin{vmatrix} 1 & 0 \\ 0 & 2 \end{vmatrix} = -(2 - 0) = -2$
$C_{33} = +\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 - 0 = 1$
The cofactor matrix is $C = \begin{bmatrix} 1 & 4 & -2 \\ -2 & 1 & 4 \\ 4 & -2 & 1 \end{bmatrix}$.
The adjoint of $A$ is the transpose of the cofactor matrix:
$adj(A) = C^T = \begin{bmatrix} 1 & -2 & 4 \\ 4 & 1 & -2 \\ -2 & 4 & 1 \end{bmatrix}$.

(C) We know that for any two square matrices $A$ and $B$ of the same order,the property of the adjoint of a product is given by:
$Adj(AB) = (Adj B)(Adj A)$.
Therefore,subtracting $(Adj B)(Adj A)$ from both sides,we get:
$Adj(AB) - (Adj B)(Adj A) = O$,where $O$ is the zero matrix of the same order.

(A) Given $A = \begin{bmatrix} 3 & 2 \\ 0 & 1 \end{bmatrix}$.
First,find the determinant: $|A| = (3)(1) - (2)(0) = 3$.
Next,find the adjoint matrix: $Adj(A) = \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}$.
Thus,$A^{-1} = \frac{1}{|A|} Adj(A) = \frac{1}{3} \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix}$.
Now,calculate $(A^{-1})^2 = \left( \frac{1}{3} \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} \right) \left( \frac{1}{3} \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} \right) = \frac{1}{9} \begin{bmatrix} 1 & -8 \\ 0 & 9 \end{bmatrix}$.
Finally,$(A^{-1})^3 = (A^{-1})^2 \cdot A^{-1} = \frac{1}{9} \begin{bmatrix} 1 & -8 \\ 0 & 9 \end{bmatrix} \cdot \frac{1}{3} \begin{bmatrix} 1 & -2 \\ 0 & 3 \end{bmatrix} = \frac{1}{27} \begin{bmatrix} 1 & -2 - 24 \\ 0 & 27 \end{bmatrix} = \frac{1}{27} \begin{bmatrix} 1 & -26 \\ 0 & 27 \end{bmatrix}$.

(B) We know that for any square matrix $A$ of order $n$,the property $A(\text{adj } A) = |A|I$ holds,where $I$ is the identity matrix of order $n$.
Given that $A$ is a $2 \times 2$ matrix,we have $A(\text{adj } A) = |A| \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} |A| & 0 \\ 0 & |A| \end{bmatrix}$.
Comparing this with the given matrix $\begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$,we get $|A| = 10$.

(A) Let $A = \begin{bmatrix} 4 & 7 \\ 1 & 2 \end{bmatrix}$.
First,calculate the determinant $|A| = (4 \times 2) - (7 \times 1) = 8 - 7 = 1$.
Next,find the adjoint matrix $Adj(A)$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$Adj(A) = \begin{bmatrix} 2 & -7 \\ -1 & 4 \end{bmatrix}$.
The inverse matrix is given by $A^{-1} = \frac{1}{|A|} Adj(A) = \frac{1}{1} \begin{bmatrix} 2 & -7 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 2 & -7 \\ -1 & 4 \end{bmatrix}$.

(C) First,we calculate the determinant of $A$:
$|A| = 3(-3 + 4) - (-3)(2 - 0) + 4(-2 - 0) = 3(1) + 3(2) + 4(-2) = 3 + 6 - 8 = 1$.
Next,we find the cofactor matrix $C$:
$C_{11} = (-3+4) = 1, C_{12} = -(2-0) = -2, C_{13} = (-2-0) = -2$
$C_{21} = -(-3+4) = -1, C_{22} = (3-0) = 3, C_{23} = -(-3-0) = 3$
$C_{31} = (-12+12) = 0, C_{32} = -(12-8) = -4, C_{33} = (-9+6) = -3$
Wait,let us recompute $A^{-1}$ using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$:
$A^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.
Now,calculate $A^2 = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix}$.
Now,calculate $A^3 = A^2 \cdot A = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.
Since $A^{-1} = A^3$,the correct option is $C$.

(B) We know that for any square matrix $A$ of order $n$,the relationship between the matrix and its adjoint is given by $A(\text{adj } A) = |A| I_n$,where $I_n$ is the identity matrix of order $n$.
Taking the determinant on both sides,we get $|A(\text{adj } A)| = ||A| I_n|$.
Using the property $|AB| = |A||B|$,we have $|A| |\text{adj } A| = |A|^n |I_n|$.
Since $|I_n| = 1$,this simplifies to $|A| |\text{adj } A| = |A|^n$.
Given that $|A| = d$,we have $d |\text{adj } A| = d^n$.
Therefore,$|\text{adj } A| = d^{n-1}$.

(B) We know that for any two non-singular square matrices $A$ and $B$ of the same order,the property of the adjoint of a product is given by:
$adj(AB) = adj(B) \cdot adj(A)$.
This is a standard property derived from the fact that $adj(AB) = |AB|(AB)^{-1} = |A||B|B^{-1}A^{-1} = (|B|B^{-1})(|A|A^{-1}) = adj(B) \cdot adj(A)$.
Therefore,the correct option is $(B)$.

(D) Let $A = \begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}$.
First,we find the determinant $|A| = 1(1(1) - 2(0)) - 2(0(1) - 2(0)) + (-3)(0(0) - 1(0)) = 1(1) - 0 + 0 = 1$.
Next,we find the cofactor matrix $C$ where $C_{ij} = (-1)^{i+j} M_{ij}$.
$C_{11} = +|\begin{smallmatrix} 1 & 2 \\ 0 & 1 \end{smallmatrix}| = 1$,$C_{12} = -|\begin{smallmatrix} 0 & 2 \\ 0 & 1 \end{smallmatrix}| = 0$,$C_{13} = +|\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}| = 0$.
$C_{21} = -|\begin{smallmatrix} 2 & -3 \\ 0 & 1 \end{smallmatrix}| = -2$,$C_{22} = +|\begin{smallmatrix} 1 & -3 \\ 0 & 1 \end{smallmatrix}| = 1$,$C_{23} = -|\begin{smallmatrix} 1 & 2 \\ 0 & 0 \end{smallmatrix}| = 0$.
$C_{31} = +|\begin{smallmatrix} 2 & -3 \\ 1 & 2 \end{smallmatrix}| = 4 - (-3) = 7$,$C_{32} = -|\begin{smallmatrix} 1 & -3 \\ 0 & 2 \end{smallmatrix}| = -(2 - 0) = -2$,$C_{33} = +|\begin{smallmatrix} 1 & 2 \\ 0 & 1 \end{smallmatrix}| = 1$.
Thus,$adj(A) = C^T = \begin{bmatrix} 1 & -2 & 7 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix}$.
Since $A^{-1} = \frac{1}{|A|} adj(A)$ and $|A| = 1$,we have $A^{-1} = \begin{bmatrix} 1 & -2 & 7 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix}$.
The element in the first row and third column is $7$.

(C) By definition,the inverse of a matrix $A$ is a matrix $B$ such that $AB = BA = I$,where $I$ is the identity matrix.
For the identity matrix ${I_3}$,we have ${I_3} \times {I_3} = {I_3}$.
Comparing this with the definition $A \times A^{-1} = I$,we can see that ${I_3}^{-1} = {I_3}$.

(B) For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,the adjoint of $A$ is given by $\text{adj } A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Given $A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$,we have $a = 1, b = -1, c = 2, d = 3$.
Substituting these values into the formula:
$\text{adj } A = \begin{bmatrix} 3 & -(-1) \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix}$.
Thus,the correct option is $B$.

(B) The inverse of a square matrix $A$ is given by the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,the determinant is $|A| = ad - bc$.
The adjoint of matrix $A$,denoted as $\text{adj}(A)$,is obtained by interchanging the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

(C) Given the equation: ${A^2} - A + I = 0$
Rearranging the terms to isolate $I$:
$I = A - {A^2}$
Factor out $A$ from the right side:
$I = A(I - A)$
Multiply both sides by ${A^{-1}}$ from the left:
${A^{-1}}I = {A^{-1}}A(I - A)$
Since ${A^{-1}}A = I$ and ${A^{-1}}I = {A^{-1}}$:
${A^{-1}} = I(I - A)$
Therefore:
${A^{-1}} = I - A$

(B) Given that $A$ and $B$ are two invertible matrices of suitable orders.
By the property of the inverse of a product of matrices,the inverse of the product of two invertible matrices is the product of their inverses in reverse order.
Therefore,$(AB)^{-1} = B^{-1}A^{-1}$.
Thus,the correct option is $(B)$.

(B) Given $A = \begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$.
First,calculate the determinant $|A| = (1)(-5) - (2)(3) = -5 - 6 = -11$.
Next,find the adjoint of $A$,denoted as $adj(A)$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$,the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Thus,$adj(A) = \begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix}$.
The inverse is given by $A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-11} \begin{bmatrix} -5 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} \frac{5}{11} & \frac{2}{11} \\ \frac{3}{11} & -\frac{1}{11} \end{bmatrix}$.

(D) Given,$A = \begin{bmatrix} 2 & 3 \\ 4 & 6 \end{bmatrix}$.
We know that the inverse of a matrix $A$ is given by ${A^{-1}} = \frac{adj(A)}{|A|}$,provided that $|A| \neq 0$.
First,we calculate the determinant of $A$:
$|A| = (2 \times 6) - (3 \times 4) = 12 - 12 = 0$.
Since the determinant of matrix $A$ is $0$,the matrix $A$ is a singular matrix.
Therefore,the inverse of $A$ does not exist.

(B) Given $A = \begin{bmatrix} 4 & 2 \\ 3 & 4 \end{bmatrix}$.
The adjoint of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Therefore,$adj\,A = \begin{bmatrix} 4 & -2 \\ -3 & 4 \end{bmatrix}$.
Now,calculate the determinant $|adj\,A|$:
$|adj\,A| = (4 \times 4) - (-3 \times -2)$
$|adj\,A| = 16 - 6$
$|adj\,A| = 10$.
Alternatively,using the property $|adj\,A| = |A|^{n-1}$ where $n$ is the order of the matrix:
$|A| = (4 \times 4) - (3 \times 2) = 16 - 6 = 10$.
Since $n=2$,$|adj\,A| = |A|^{2-1} = |A|^1 = 10$.

(B) Let $A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$.
The cofactor $C_{ij}$ of an element $a_{ij}$ is given by $(-1)^{i+j} M_{ij}$,where $M_{ij}$ is the minor.
$C_{11} = + \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} = (-3)(1) - (4)(-1) = -3 + 4 = 1$
$C_{12} = - \begin{vmatrix} 2 & 4 \\ 0 & 1 \end{vmatrix} = -(2 - 0) = -2$
$C_{13} = + \begin{vmatrix} 2 & -3 \\ 0 & -1 \end{vmatrix} = (-2 - 0) = -2$
$C_{21} = - \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} = -(-3 + 4) = -1$
$C_{22} = + \begin{vmatrix} 3 & 4 \\ 0 & 1 \end{vmatrix} = (3 - 0) = 3$
$C_{23} = - \begin{vmatrix} 3 & -3 \\ 0 & -1 \end{vmatrix} = -(-3 - 0) = 3$
$C_{31} = + \begin{vmatrix} -3 & 4 \\ -3 & 4 \end{vmatrix} = (-12 + 12) = 0$
$C_{32} = - \begin{vmatrix} 3 & 4 \\ 2 & 4 \end{vmatrix} = -(12 - 8) = -4$
$C_{33} = + \begin{vmatrix} 3 & -3 \\ 2 & -3 \end{vmatrix} = (-9 + 6) = -3$
The adjoint matrix is the transpose of the cofactor matrix:
$adj(A) = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.

(A) Given $A = \begin{bmatrix} 0 & 3 \\ 2 & 0 \end{bmatrix}$.
First,we find the determinant of $A$:
$|A| = (0 \times 0) - (3 \times 2) = 0 - 6 = -6$.
Next,we find the adjoint of $A$:
$adj(A) = \begin{bmatrix} 0 & -3 \\ -2 & 0 \end{bmatrix}$.
We know that $A^{-1} = \frac{1}{|A|} adj(A)$.
Substituting the values,we get:
$A^{-1} = \frac{1}{-6} adj(A) = \frac{-1}{6} adj(A)$.
Comparing this with $A^{-1} = \lambda (adj(A))$,we find that $\lambda = \frac{-1}{6}$.

(B) For any two invertible matrices $A$ and $B$ of the same order,the inverse of their product is given by the reversal law of inverses.
We know that $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = A(I)A^{-1} = AA^{-1} = I$.
Similarly,$(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}(I)B = B^{-1}B = I$.
Since $(AB)(B^{-1}A^{-1}) = I$ and $(B^{-1}A^{-1})(AB) = I$,it follows that $(AB)^{-1} = B^{-1}A^{-1}$.
Therefore,option $B$ is correct.

(A) We know that for any square matrix $A$,the product of the matrix and its adjoint is given by the property: $A \cdot (adj(A)) = |A| \cdot I$,where $I$ is the identity matrix of the same order.
First,we calculate the determinant of matrix $A$:
$|A| = \begin{vmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{vmatrix} = (\cos x)(\cos x) - (\sin x)(-\sin x) = \cos^2 x + \sin^2 x = 1$.
Since $|A| = 1$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,we have:
$A \cdot (adj(A)) = 1 \cdot \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.