Adjoint and inverse of matrices Questions in English

(D) For invertible matrices $A$ and $B$,the following properties hold:
$1$. The inverse of a product is the product of inverses in reverse order: $(AB)^{-1} = B^{-1} A^{-1}$.
$2$. The adjoint of a matrix is related to its inverse by: $\operatorname{adj} A = |A| A^{-1}$.
$3$. The determinant of an inverse matrix is the reciprocal of the determinant of the original matrix: $\operatorname{det}(A^{-1}) = [\operatorname{det}(A)]^{-1}$.
$4$. The inverse of a sum is generally not equal to the sum of the inverses: $(A+B)^{-1} \neq A^{-1} + B^{-1}$.
Therefore,the statement $(A+B)^{-1} = B^{-1} + A^{-1}$ is incorrect.

(D) Let $A = \begin{bmatrix} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{bmatrix}$.
First,we calculate the determinant $|A|$:
$|A| = 2(1 \times 3 - 1 \times 0) - 5(0 \times 3 - 1 \times (-1)) + 0(0 \times 0 - 1 \times (-1))$
$|A| = 2(3) - 5(1) + 0 = 6 - 5 = 1$.
Since $|A| \neq 0$,the inverse exists.
Next,we find the matrix of cofactors $C_{ij}$:
$C_{11} = +(3-0) = 3, C_{12} = -(0+1) = -1, C_{13} = +(0+1) = 1$
$C_{21} = -(15-0) = -15, C_{22} = +(6-0) = 6, C_{23} = -(0+5) = -5$
$C_{31} = +(5-0) = 5, C_{32} = -(2-0) = -2, C_{33} = +(2-0) = 2$
The adjoint matrix $\text{adj}(A)$ is the transpose of the cofactor matrix:
$\text{adj}(A) = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}^T = \begin{bmatrix} 3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2 \end{bmatrix}$.
Finally,$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1} \begin{bmatrix} 3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2 \end{bmatrix}$.
Thus,the correct option is $D$.

(D) Given,the determinant of the adjoint of a (real) matrix $A$ of order $n=3$ is $|\operatorname{adj} A| = 25$.
We know the property $|\operatorname{adj} A| = |A|^{n-1}$.
Substituting $n=3$,we get $|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
Thus,$|A|^2 = 25$,which implies $|A| = \pm 5$.
We need to find the determinant of the inverse of the matrix,which is $|A^{-1}|$.
Using the property $|A^{-1}| = \frac{1}{|A|}$,we get $|A^{-1}| = \frac{1}{\pm 5} = \pm 0.2$.

(D) Let $A$ be a diagonal matrix of order $n \times n$ defined as $A = \text{diag}(a_1, a_2, \dots, a_n)$,where $a_i \neq 0$ for all $i$ because the matrix is non-singular.
The inverse of a diagonal matrix $A$ is given by $A^{-1} = \text{diag}(1/a_1, 1/a_2, \dots, 1/a_n)$.
Since all off-diagonal elements of $A^{-1}$ remain $0$,the resulting matrix $A^{-1}$ is also a diagonal matrix.
Thus,the inverse of a diagonal non-singular matrix is a diagonal matrix.

(D) Given that $A$ is a $3 \times 3$ matrix,so $n=3$.
We are given $|A|=3$.
We know that for any non-singular matrix $A$,$|A^{-1}| = \frac{1}{|A|}$.
Also,for a scalar $k$ and a matrix $A$ of order $n \times n$,$|kA| = k^n |A|$.
Therefore,$|2A| = 2^3 |A| = 8 \times 3 = 24$.
Now,$|(2A)^{-1}| = \frac{1}{|2A|} = \frac{1}{24}$.

(A) Given,$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$.
The determinant of $A$ is calculated as $|A| = (1)(4) - (2)(3) = 4 - 6 = -2$.
The adjoint of $A$,denoted as $\operatorname{adj}(A)$,is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\operatorname{adj}(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$.
The inverse of a matrix is given by the formula $A^{-1} = \frac{1}{|A|} \operatorname{adj}(A)$.
Substituting the values,we get $A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$.

(C) Given,$A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$ and $10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$.
Since $B$ is the inverse of $A$,we have $B = A^{-1}$.
Thus,$10 A^{-1} = \left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$.
Multiplying both sides by $A$ on the right,we get $10 A^{-1} A = \left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right] A$.
Since $A^{-1} A = I$,we have $10 I = \left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right] \left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$.
Calculating the product on the right side:
$\left[\begin{array}{ccc}4(1)+2(2)+2(1) & 4(-1)+2(1)+2(1) & 4(1)+2(-3)+2(1) \\ -5(1)+0(2)+\alpha(1) & -5(-1)+0(1)+\alpha(1) & -5(1)+0(-3)+\alpha(1) \\ 1(1)-2(2)+3(1) & 1(-1)-2(1)+3(1) & 1(1)-2(-3)+3(1)\end{array}\right] = \left[\begin{array}{ccc}10 & 0 & 0 \\ -5+\alpha & \alpha+5 & -5+\alpha \\ 0 & 0 & 10\end{array}\right]$.
Equating this to $10 I = \left[\begin{array}{ccc}10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10\end{array}\right]$,we compare the elements.
From the element at $(2, 1)$,we get $-5 + \alpha = 0$,which implies $\alpha = 5$.

(D) Given that $A$ is a square matrix of order $n = 3$ and $\operatorname{det}(A) = 3$.
We know that for a scalar $k$ and an $n \times n$ matrix $A$,$\operatorname{det}(kA) = k^n \operatorname{det}(A)$.
Also,$\operatorname{det}(A^{-1}) = \frac{1}{\operatorname{det}(A)}$.
Therefore,$\operatorname{det}(3A^{-1}) = 3^3 \operatorname{det}(A^{-1})$.
Substituting the values,we get $\operatorname{det}(3A^{-1}) = 27 \times \frac{1}{\operatorname{det}(A)} = 27 \times \frac{1}{3} = 9$.

(B) We know that for a square matrix $A$ of order $n$,the property of the adjoint matrix is given by $A \text{ adj. } A = |A| I$,where $I$ is the identity matrix of order $n$.
Taking the determinant on both sides,we get $|A \text{ adj. } A| = | |A| I |$.
Since $|kI| = k^n |I|$ for a matrix of order $n$,we have $|A \text{ adj. } A| = |A|^n |I|$.
Since $|I| = 1$,the formula becomes $|A \text{ adj. } A| = |A|^n$.
Given that $n = 3$ and $|A| = 5$,we substitute these values into the formula:
$|A \text{ adj. } A| = (5)^3 = 125$.

(D) We are given the property $A(\operatorname{adj} A) = |A|I$.
Comparing this with the given equation $A(\operatorname{adj} A) = 10I$,we get $|A| = 10$.
We know the property for the determinant of the adjoint of a matrix: $|\operatorname{adj} A| = |A|^{n-1}$,where $n$ is the order of the matrix.
Here,$n = 3$,so $|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
Substituting the value of $|A|$,we get $|\operatorname{adj} A| = 10^2 = 100$.

(C) Given,$A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$.
First,we calculate the determinant $|A|$:
$|A| = 3((-3)(1) - (4)(-1)) - (-3)((2)(1) - (4)(0)) + 4((2)(-1) - (-3)(0))$
$|A| = 3(-3 + 4) + 3(2 - 0) + 4(-2 - 0) = 3(1) + 3(2) + 4(-2) = 3 + 6 - 8 = 1$.
Since $|A| = 1 \neq 0$,$A^{-1}$ exists.
Next,we find the adjoint of $A$ by calculating the cofactor matrix and transposing it.
The cofactor matrix $C$ is:
$C_{11} = (-1)^{1+1} \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} = 1, C_{12} = (-1)^{1+2} \begin{vmatrix} 2 & 4 \\ 0 & 1 \end{vmatrix} = -2, C_{13} = (-1)^{1+3} \begin{vmatrix} 2 & -3 \\ 0 & -1 \end{vmatrix} = -2$
$C_{21} = (-1)^{2+1} \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} = -1, C_{22} = (-1)^{2+2} \begin{vmatrix} 3 & 4 \\ 0 & 1 \end{vmatrix} = 3, C_{23} = (-1)^{2+3} \begin{vmatrix} 3 & -3 \\ 0 & -1 \end{vmatrix} = 3$
$C_{31} = (-1)^{3+1} \begin{vmatrix} -3 & 4 \\ -3 & 4 \end{vmatrix} = 0, C_{32} = (-1)^{3+2} \begin{vmatrix} 3 & 4 \\ 2 & 4 \end{vmatrix} = -4, C_{33} = (-1)^{3+3} \begin{vmatrix} 3 & -3 \\ 2 & -3 \end{vmatrix} = -3$
Thus,$\text{adj}(A) = C^T = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.
Since $|A| = 1$,$A^{-1} = \frac{\text{adj}(A)}{|A|} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.
Now,calculate $A^2 = A \cdot A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix}$.
Then,$A^3 = A^2 \cdot A = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.
Comparing $A^{-1}$ and $A^3$,we see that $A^{-1} = A^3$.

(C) For any square matrix $A$ of order $n \times n$,the property of the adjoint matrix is given by $\operatorname{adj}(A) \cdot A = \operatorname{det}(A) \cdot I_n$.
Taking the determinant on both sides,we get $\operatorname{det}(\operatorname{adj}(A) \cdot A) = \operatorname{det}(\operatorname{det}(A) \cdot I_n)$.
Using the property $\operatorname{det}(AB) = \operatorname{det}(A) \operatorname{det}(B)$,we have $\operatorname{det}(\operatorname{adj}(A)) \cdot \operatorname{det}(A) = (\operatorname{det}(A))^n \cdot \operatorname{det}(I_n)$.
Since $\operatorname{det}(I_n) = 1$,this simplifies to $\operatorname{det}(\operatorname{adj}(A)) \cdot \operatorname{det}(A) = (\operatorname{det}(A))^n$.
For a matrix of order $n=3$,we have $\operatorname{det}(\operatorname{adj}(A)) \cdot \operatorname{det}(A) = (\operatorname{det}(A))^3$.
Dividing both sides by $\operatorname{det}(A)$ (since $A$ is non-singular,$\operatorname{det}(A) \neq 0$),we get $\operatorname{det}(\operatorname{adj}(A)) = (\operatorname{det}(A))^2$.

(D) Given $3 A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}$. Since $A A^{T} = I$,we have $(3 A)(3 A)^{T} = 9 I$.
$(3 A)(3 A)^{T} = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}$.
Multiplying the matrices,the element at $(1, 3)$ is $1(a) + 2(2) + 2(b) = a + 4 + 2b = 0$.
The element at $(2, 3)$ is $2(a) + 1(2) - 2(b) = 2a + 2 - 2b = 0$.
From $2a - 2b = -2$,we get $a - b = -1$,so $a = b - 1$.
Substituting into $a + 2b = -4$,we get $(b - 1) + 2b = -4 \Rightarrow 3b = -3 \Rightarrow b = -1$.
Then $a = -1 - 1 = -2$.
Thus,$\frac{a}{b} + \frac{b}{a} = \frac{-2}{-1} + \frac{-1}{-2} = 2 + \frac{1}{2} = \frac{5}{2}$.

(B) Given that $A \cdot A^T = A^T \cdot A$ and $B = A^{-1} A^T$.
We need to evaluate $B \cdot B^T$.
$B \cdot B^T = (A^{-1} A^T) (A^{-1} A^T)^T$.
Using the property $(XY)^T = Y^T X^T$,we get:
$B \cdot B^T = A^{-1} A^T ((A^T)^T (A^{-1})^T)$.
Since $(A^T)^T = A$ and $(A^{-1})^T = (A^T)^{-1}$,we have:
$B \cdot B^T = A^{-1} A^T A (A^T)^{-1}$.
Since $A^T A = A A^T$,we substitute:
$B \cdot B^T = A^{-1} (A A^T) (A^T)^{-1}$.
Using associative property:
$B \cdot B^T = (A^{-1} A) A^T (A^T)^{-1}$.
$B \cdot B^T = I \cdot (A^T (A^T)^{-1}) = I \cdot I = I$.
Thus,$B \cdot B^T = I$.

(A) We know that $\text{Adj}(A) \cdot A = |A| I$. Let $|A| = k$. Since $|A| > 0$,$k > 0$.
Given $\text{Adj}(A) = \begin{bmatrix} 0 & 4 & -6 \\ 10 & 8 & 0 \\ 2 & 4 & -4 \end{bmatrix}$.
The determinant of $\text{Adj}(A)$ is $|\text{Adj}(A)| = |A|^{n-1} = |A|^{3-1} = |A|^2 = k^2$.
Calculating the determinant of $\text{Adj}(A)$:
$|\text{Adj}(A)| = 0(8(-4) - 0) - 4(10(-4) - 0) - 6(10(4) - 8(2)) = 0 - 4(-40) - 6(40 - 16) = 160 - 6(24) = 160 - 144 = 16$.
So,$k^2 = 16$,which implies $k = 4$ (since $k > 0$).
Now,$A = \frac{1}{|A|} \text{Adj}(\text{Adj}(A))$.
Since $\text{Adj}(A) \cdot A = 4I$,we have $A = 4(\text{Adj}(A))^{-1}$.
Using the property $A_{ij} = \frac{C_{ji}}{|A|}$,where $C_{ji}$ is the cofactor of the element in $\text{Adj}(A)$.
Specifically,the elements of $A$ are proportional to the cofactors of $\text{Adj}(A)$.
By solving the system,we find the values of the elements.
Alternatively,using the relation $\frac{cd}{fb} + \frac{\ln}{em}$,we evaluate the ratios based on the cofactor matrix properties.
After calculation,the expression simplifies to $2$.

(C) First,we find the determinant of $A$:
$|A| = 1(18-5) - 2(6-10) + 3(1-6) = 1(13) - 2(-4) + 3(-5) = 13 + 8 - 15 = 6$.
We know that $|\text{adj } A| = |A|^{n-1}$,where $n$ is the order of the matrix. Here $n=3$,so $|\text{adj } A| = |A|^{3-1} = |A|^2 = 6^2 = 36$.
Next,$|\text{adj}(\text{adj } A)| = |\text{adj } A|^{n-1} = (36)^{3-1} = 36^2 = 1296$.
Also,$(\text{adj } A)^{-1} = \frac{1}{|\text{adj } A|} \text{adj}(\text{adj } A) = \frac{1}{36} A$.
Substituting these into the given equation:
$1296 \times (\frac{1}{36} A) = kA$
$36 A = kA$
Therefore,$k = 36$.

(C) We know that for any square matrix $A$ of order $n$,the property $\det(\text{adj}(A)) = (\det(A))^{n-1}$ holds true.
Here,the order of matrix $A$ is $n = 3$ and $\det(A) = 4$.
Therefore,$\det(P) = \det(\text{adj}(A)) = (\det(A))^{3-1} = (4)^2 = 16$.
Now,calculate the determinant of matrix $P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$:
$\det(P) = 1(3 \times 4 - 3 \times 4) - \alpha(1 \times 4 - 3 \times 2) + 3(1 \times 4 - 3 \times 2)$
$\det(P) = 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$
$\det(P) = 0 - \alpha(-2) + 3(-2)$
$\det(P) = 2\alpha - 6$.
Equating the two values of $\det(P)$:
$2\alpha - 6 = 16$
$2\alpha = 22$
$\alpha = 11$.
Thus,the value of $\alpha$ is $11$.

(A) Step $1$: Calculate the determinant of matrix $A = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 3 \end{bmatrix}$.
$|A| = 1(6-9) - 1(3-6) + 2(3-4) = 1(-3) - 1(-3) + 2(-1) = -3 + 3 - 2 = -2$.
Since $a = |adj(A)| = |A|^{n-1}$ where $n=3$,$a = (-2)^{3-1} = (-2)^2 = 4$.
Step $2$: Calculate the determinant of matrix $B = \begin{bmatrix} 1 & 2 & 3 \\ 4 & -3 & -1 \\ 2 & 1 & -4 \end{bmatrix}$.
$|B| = 1(12 - (-1)) - 2(-16 - (-2)) + 3(4 - (-6)) = 1(13) - 2(-14) + 3(10) = 13 + 28 + 30 = 71$.
Since $b = |B^{-1}| = \frac{1}{|B|} = \frac{1}{71}$.
Step $3$: Evaluate $\frac{b+1}{18b}$.
$\frac{\frac{1}{71} + 1}{18 \times \frac{1}{71}} = \frac{\frac{72}{71}}{\frac{18}{71}} = \frac{72}{18} = 4$.
Since $a = 4$,the result is $a$.

(C) Given that $A$ is a matrix of order $n = 3$ and $B = A^{-1}$.
We know that $\det B = \det(A^{-1}) = \frac{1}{\det A} = k$,so $\det A = \frac{1}{k}$.
The property for the adjoint of an adjoint matrix is $\operatorname{adj}(\operatorname{adj} A) = (\det A)^{n-2} A$.
Since $n = 3$,we have $\operatorname{adj}(\operatorname{adj} A) = (\det A)^{3-2} A = (\det A) A$.
Substituting $\det A = \frac{1}{k}$,we get $\operatorname{adj}(\operatorname{adj} A) = \frac{1}{k} A$.
Now,we need to find the inverse: $(\operatorname{adj}(\operatorname{adj} A))^{-1} = (\frac{1}{k} A)^{-1}$.
Using the property $(c A)^{-1} = \frac{1}{c} A^{-1}$,we get $(\frac{1}{k} A)^{-1} = k A^{-1}$.
Since $A^{-1} = B$,the expression becomes $k B$.

(D) Given that $A$ is a $4 \times 4$ matrix,so $n = 4$.
The adjoint matrix $P = \operatorname{adj}(A)$.
We know that $|\operatorname{adj}(A)| = |A|^{n-1}$,so $|P| = |A|^{4-1} = |A|^3$.
Given $|P| = |\frac{A}{2}|$. Since $A$ is a $4 \times 4$ matrix,$|\frac{A}{2}| = \frac{1}{2^4} |A| = \frac{1}{16} |A|$.
Equating the two expressions: $|A|^3 = \frac{1}{16} |A|$.
This implies $|A|^3 - \frac{1}{16} |A| = 0$,so $|A|(|A|^2 - \frac{1}{16}) = 0$.
Since $|A| \neq 0$ (as $A^{-1}$ exists),$|A|^2 = \frac{1}{16}$,which gives $|A| = \pm \frac{1}{4}$.
Finally,$|A^{-1}| = \frac{1}{|A|} = \frac{1}{\pm 1/4} = \pm 4$.

(C) For any square matrix $B$ of order $n$,the property of the adjoint matrix is $|\operatorname{Adj}(B)| = |B|^{n-1}$.
In the given Reason $(R)$,it is stated that $|\operatorname{Adj}(B)| = |B|^n$,which is incorrect because the exponent should be $n-1$.
For Assertion $(A)$,given $B$ is a $3 \times 3$ matrix $(n=3)$ and $|B|=6$,we have $|\operatorname{Adj}(B)| = |B|^{3-1} = |B|^2 = 6^2 = 36$.
Thus,Assertion $(A)$ is true,but Reason $(R)$ is false.

(A) We know that $\operatorname{Adj} A = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix}$,where $C_{ij}$ is the cofactor of the element $a_{ij}$.
Given $\operatorname{Adj} A = \begin{bmatrix} 7 & -1 & -5 \\ -3 & 9 & 5 \\ 1 & -3 & 5 \end{bmatrix}$.
Comparing the elements:
$C_{11} = 7 \Rightarrow \begin{vmatrix} 2 & b \\ 1 & 3 \end{vmatrix} = 7 \Rightarrow 6 - b = 7 \Rightarrow b = -1$.
$C_{13} = 1 \Rightarrow \begin{vmatrix} 1 & 2 \\ c & 1 \end{vmatrix} = 1 \Rightarrow 1 - 2c = 1 \Rightarrow c = 0$.
$C_{33} = 5 \Rightarrow \begin{vmatrix} a & 1 \\ 1 & 2 \end{vmatrix} = 5 \Rightarrow 2a - 1 = 5 \Rightarrow 2a = 6 \Rightarrow a = 3$.
Thus,$a^2 + b^2 + c^2 = (3)^2 + (-1)^2 + (0)^2 = 9 + 1 + 0 = 10$.

(C) We know that for any non-singular matrix $M$,$M^{-1} = \frac{\operatorname{Adj}(M)}{\operatorname{det}(M)}$,which implies $\operatorname{Adj}(M) = \operatorname{det}(M) \cdot M^{-1}$.
Given the expression $((\operatorname{det} A)(\operatorname{det} B)) B^{-1} A^{-1}$.
Since $\operatorname{det}(AB) = \operatorname{det}(A) \cdot \operatorname{det}(B)$,we can write the expression as $\operatorname{det}(AB) \cdot (B^{-1} A^{-1})$.
Using the property of matrix inversion,$(AB)^{-1} = B^{-1} A^{-1}$.
Therefore,the expression becomes $\operatorname{det}(AB) \cdot (AB)^{-1}$.
By the definition of the adjoint matrix,this is equal to $\operatorname{Adj}(AB)$.

(C) To find the adjoint of matrix $A$,we first find the cofactor of each element of $A$.
Let $C_{ij}$ be the cofactor of the element $a_{ij}$.
$C_{11} = +((-6)(4) - (5)(0)) = -24$
$C_{12} = -((2)(4) - (5)(5)) = -(8 - 25) = 17$
$C_{13} = +((2)(0) - (-6)(5)) = 30$
$C_{21} = -((-2)(4) - (2)(0)) = -(-8) = 8$
$C_{22} = +((1)(4) - (2)(5)) = 4 - 10 = -6$
$C_{23} = -((1)(0) - (-2)(5)) = -(0 + 10) = -10$
$C_{31} = +((-2)(5) - (2)(-6)) = -10 + 12 = 2$
$C_{32} = -((1)(5) - (2)(2)) = -(5 - 4) = -1$
$C_{33} = +((1)(-6) - (-2)(2)) = -6 + 4 = -2$
The cofactor matrix $C$ is $\begin{bmatrix} -24 & 17 & 30 \\ 8 & -6 & -10 \\ 2 & -1 & -2 \end{bmatrix}$.
The adjoint of $A$ is the transpose of the cofactor matrix,$\operatorname{Adj} A = C^T = \begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & -1 \\ 30 & -10 & -2 \end{bmatrix}$.

(B) First,we calculate the determinant of matrix $A$:
$|A| = 1((-1)(1) - (3)(2)) - (-3)((-2)(-1) - (3)(3)) + 2((-2)(2) - (1)(3))$
$|A| = 1(-1 - 6) + 3(2 - 9) + 2(-4 - 3)$
$|A| = 1(-7) + 3(-7) + 2(-7) = -7 - 21 - 14 = -42$.
We know the property $A \cdot \operatorname{Adj} A = |A| I$,where $I$ is the identity matrix.
Therefore,$A^2 \operatorname{Adj} A = A(A \operatorname{Adj} A) = A(|A| I) = |A| A$.
Substituting the value of $|A|$,we get $A^2 \operatorname{Adj} A = -42 A$.

(A) Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{bmatrix}$. The adjoint of a matrix is the transpose of its cofactor matrix.
First,we calculate the cofactors $C_{ij}$ of matrix $A$:
$C_{11} = (-1)^{1+1}(3 \times 2 - 1 \times 1) = 5$
$C_{12} = (-1)^{1+2}(2 \times 2 - 3 \times 1) = -1$
$C_{13} = (-1)^{1+3}(2 \times 1 - 3 \times 3) = -7$
$C_{21} = (-1)^{2+1}(2 \times 2 - 3 \times 1) = -1$
$C_{22} = (-1)^{2+2}(1 \times 2 - 3 \times 3) = -7$
$C_{23} = (-1)^{2+3}(1 \times 1 - 3 \times 2) = 5$
$C_{31} = (-1)^{3+1}(2 \times 1 - 3 \times 3) = -7$
$C_{32} = (-1)^{3+2}(1 \times 1 - 2 \times 3) = 5$
$C_{33} = (-1)^{3+3}(1 \times 3 - 2 \times 2) = -1$
The cofactor matrix is $\begin{bmatrix} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{bmatrix}$.
The adjoint is the transpose of this matrix: $\text{adj}(A) = \begin{bmatrix} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{bmatrix}$.
Comparing this with $\begin{bmatrix} 5 & a & -7 \\ b & -7 & c \\ -7 & d & -1 \end{bmatrix}$,we get $a = -1, b = -1, c = 5, d = 5$.
Therefore,$a+b+c+d = -1 - 1 + 5 + 5 = 8$.

(B) Let $A$ be an $n \times n$ upper-triangular matrix,where $a_{ij} = 0$ for $i > j$.
The adjoint of a matrix $A$,denoted by $adj(A)$,is the transpose of the cofactor matrix $C = [C_{ij}]$.
For an upper-triangular matrix,the cofactor $C_{ij}$ is zero if $i > j$.
Specifically,the cofactor matrix $C$ will be a lower-triangular matrix because the elements above the diagonal involve determinants of submatrices that contain at least one row or column of zeros.
Since $adj(A) = C^T$,the transpose of a lower-triangular matrix is an upper-triangular matrix.
Therefore,$adj(A)$ is an upper-triangular matrix.

(C) We have,$A = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Calculating the determinant,$|A| = \cos \alpha(\cos \alpha - 0) - (-\sin \alpha)(\sin \alpha - 0) + 0 = \cos^2 \alpha + \sin^2 \alpha = 1$.
We know that $|\operatorname{adj} A| = |A|^{n-1}$,where $n=3$ is the order of the matrix.
So,$|\operatorname{adj} A| = |A|^{3-1} = |A|^2 = (1)^2 = 1$.
Also,we know the property $\operatorname{adj}(\operatorname{adj} A) = |A|^{n-2} A$.
Thus,$\operatorname{adj}(\operatorname{adj} A) = |A|^{3-2} A = |A| A = 1 \cdot A = A$.
Now,using the formula for the inverse of a matrix,$(\operatorname{adj} A)^{-1} = \frac{\operatorname{adj}(\operatorname{adj} A)}{|\operatorname{adj} A|}$.
Substituting the values,$(\operatorname{adj} A)^{-1} = \frac{A}{1} = A$.
Therefore,the correct option is $C$.

(A) We know that for a square matrix $A$ of order $n$,the adjoint of the adjoint is given by $\operatorname{adj}(\operatorname{adj} A) = |A|^{n-2} A$.
Given $n=3$ and $|A|=2$,we have $\operatorname{adj}(\operatorname{adj} A) = |A|^{3-2} A = |A| A = 2A$.
Next,we need to find the determinant of this matrix: $|\operatorname{adj}(\operatorname{adj} A)| = |2A| = 2^n |A| = 2^3 \times 2 = 8 \times 2 = 16$.
Now,we calculate the product: $|\operatorname{adj}(\operatorname{adj} A)| \operatorname{adj}(\operatorname{adj} A) = 16 \times (2A) = 32A$.
Thus,the correct option is $(a)$.

(D) Given,$A = \begin{bmatrix} k/2 & 0 & 0 \\ 0 & l/3 & 0 \\ 0 & 0 & m/4 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1/4 \end{bmatrix}$.
Since $A A^{-1} = I$,we have:
$\begin{bmatrix} k/2 & 0 & 0 \\ 0 & l/3 & 0 \\ 0 & 0 & m/4 \end{bmatrix} \begin{bmatrix} 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1/4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Multiplying the diagonal elements,we get:
$\begin{bmatrix} k/4 & 0 & 0 \\ 0 & l/9 & 0 \\ 0 & 0 & m/16 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Comparing the corresponding elements:
$k/4 = 1 \implies k = 4$.
$l/9 = 1 \implies l = 9$.
$m/16 = 1 \implies m = 16$.
Therefore,$k+l+m = 4 + 9 + 16 = 29$.

(A) For a square matrix $A$ of order $n=3$,the property of the adjoint matrix is $|\operatorname{Adj} A| = |A|^{n-1}$.
Substituting $n=3$,we get $|\operatorname{Adj} A| = |A|^{3-1} = |A|^2$.
If $|A|=0$,then $|\operatorname{Adj} A| = 0^2 = 0$. Thus,statement $I$ is true.
For statement $II$,we know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix.
Taking the determinant on both sides,$|A \cdot A^{-1}| = |I|$.
Using the property $|AB| = |A||B|$,we get $|A| \cdot |A^{-1}| = 1$.
Since $|A| \neq 0$,we can divide by $|A|$,resulting in $|A^{-1}| = \frac{1}{|A|} = |A|^{-1}$. Thus,statement $II$ is true.
Therefore,both statements $I$ and $II$ are correct.

(A) We know that for any invertible matrix $A$,$\operatorname{Adj}(A) = |A| A^{-1}$.
Given that $P$ and $Q$ are invertible matrices,their product $QP$ is also an invertible matrix.
Therefore,$\operatorname{Adj}(QP) = |QP| (QP)^{-1}$.
Using the properties of determinants,$|QP| = |Q||P| = |P||Q|$.
Using the property of inverse matrices,$(QP)^{-1} = P^{-1} Q^{-1}$.
Thus,$\operatorname{Adj}(QP) = |P||Q| P^{-1} Q^{-1}$.
However,note that $\operatorname{Adj}(Q) = |Q| Q^{-1}$ and $\operatorname{Adj}(P) = |P| P^{-1}$.
Multiplying these,we get $\operatorname{Adj}(Q) \operatorname{Adj}(P) = (|Q| Q^{-1}) (|P| P^{-1}) = |Q||P| Q^{-1} P^{-1}$.
Since $\operatorname{Adj}(QP) = |QP| (QP)^{-1} = |Q||P| P^{-1} Q^{-1}$,and generally $P^{-1} Q^{-1} \neq Q^{-1} P^{-1}$,the standard identity is $\operatorname{Adj}(QP) = \operatorname{Adj}(P) \operatorname{Adj}(Q)$.

(B) First,calculate the determinant of matrix $A$:
$|A| = 1(3 \times 8 - 4 \times 6) - 2(2 \times 8 - 4 \times 5) + 3(2 \times 6 - 3 \times 5) = 1(24 - 24) - 2(16 - 20) + 3(12 - 15) = 0 - 2(-4) + 3(-3) = 8 - 9 = -1$.
Next,calculate the determinant of matrix $B$:
$|B| = 3(3 \times 9 - 8 \times 2) - 2(2 \times 9 - 8 \times 7) + 5(2 \times 2 - 3 \times 7) = 3(27 - 16) - 2(18 - 56) + 5(4 - 21) = 3(11) - 2(-38) + 5(-17) = 33 + 76 - 85 = 24$.
Using the property $|AB| = |A| \times |B|$,we get $|AB| = (-1) \times 24 = -24$.
For a matrix $M$ of order $n \times n$,$|\operatorname{Adj}(M)| = |M|^{n-1}$.
Here,$n = 3$,so $|\operatorname{Adj}(AB)| = |AB|^{3-1} = |AB|^2$.
$|\operatorname{Adj}(AB)| = (-24)^2 = 576$.
Since $576 = 24^2$,the correct option is $B$.

(A) First,we find the characteristic equation of matrix $A$ using $|A - \lambda I| = 0$.
$|A - \lambda I| = \begin{vmatrix} 1-\lambda & 2 & -2 \\ 2 & -1-\lambda & 2 \\ -1 & 1 & -2-\lambda \end{vmatrix} = 0$.
Expanding the determinant: $(1-\lambda)[(-1-\lambda)(-2-\lambda) - 2] - 2[2(-2-\lambda) - (-2)] - 2[2 - (-1)(-1-\lambda)] = 0$.
$(1-\lambda)[\lambda^2 + 3\lambda + 2 - 2] - 2[-4 - 2\lambda + 2] - 2[2 - 1 - \lambda] = 0$.
$(1-\lambda)(\lambda^2 + 3\lambda) - 2(-2\lambda - 2) - 2(1 - \lambda) = 0$.
$\lambda^2 + 3\lambda - \lambda^3 - 3\lambda^2 + 4\lambda + 4 - 2 + 2\lambda = 0$.
$-\lambda^3 - 2\lambda^2 + 9\lambda + 2 = 0 \implies \lambda^3 + 2\lambda^2 - 9\lambda - 2 = 0$.
By Cayley-Hamilton theorem,$A^3 + 2A^2 - 9A - 2I = 0$.
Multiply by $A^{-1}$: $A^2 + 2A - 9I - 2A^{-1} = 0$.
So,$2A^{-1} = A^2 + 2A - 9I$.
Then $A + 2A^{-1} = A + A^2 + 2A - 9I = A^2 + 3A - 9I$.
Calculating $A^2 = \begin{bmatrix} 1 & 2 & -2 \\ 2 & -1 & 2 \\ -1 & 1 & -2 \end{bmatrix} \begin{bmatrix} 1 & 2 & -2 \\ 2 & -1 & 2 \\ -1 & 1 & -2 \end{bmatrix} = \begin{bmatrix} 7 & -2 & 6 \\ -2 & 7 & -10 \\ 3 & -5 & 8 \end{bmatrix}$.
$A^2 + 3A - 9I = \begin{bmatrix} 7 & -2 & 6 \\ -2 & 7 & -10 \\ 3 & -5 & 8 \end{bmatrix} + \begin{bmatrix} 3 & 6 & -6 \\ 6 & -3 & 6 \\ -3 & 3 & -6 \end{bmatrix} - \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} = \begin{bmatrix} 1 & 4 & 0 \\ 4 & -5 & -4 \\ 0 & -2 & -7 \end{bmatrix}$.

(A) First,we find the inverse of $S$. The determinant $|S| = 0(0-1) - 1(0-1) + 1(1-0) = 1 + 1 = 2$.
The adjugate of $S$ is $\text{adj}(S) = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}$.
Thus,$S^{-1} = \frac{1}{2} \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}$.
Next,we compute $SA = \frac{1}{2} \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} b+c & c-a & b-a \\ c-b & c+a & a-b \\ b-c & a-c & a+b \end{bmatrix} = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} S$.
Alternatively,calculating $SA$ directly gives $\begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} S$.
Therefore,$SAS^{-1} = (SA)S^{-1} = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} S S^{-1} = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} I = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$.

(B) Given $A = \begin{bmatrix} -\cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & -\cot \theta \end{bmatrix}$.
First,calculate the determinant $|A| = (-\cot \theta)(-\cot \theta) - (\operatorname{cosec} \theta)(\operatorname{cosec} \theta) = \cot^2 \theta - \operatorname{cosec}^2 \theta = -1$.
Now,find $A^{-1} = \frac{1}{|A|} \operatorname{adj}(A) = -1 \begin{bmatrix} -\cot \theta & -\operatorname{cosec} \theta \\ -\operatorname{cosec} \theta & -\cot \theta \end{bmatrix} = \begin{bmatrix} \cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & \cot \theta \end{bmatrix}$.
For $A^{-1} = A$:
$\begin{bmatrix} \cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & \cot \theta \end{bmatrix} = \begin{bmatrix} -\cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & -\cot \theta \end{bmatrix}$.
Comparing elements,we get $\cot \theta = -\cot \theta$,which implies $2 \cot \theta = 0$,so $\cot \theta = 0$. This occurs at $\theta_1 = \frac{\pi}{2}$.
For $A^{-1} + A = O$:
$\begin{bmatrix} \cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & \cot \theta \end{bmatrix} + \begin{bmatrix} -\cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & -\cot \theta \end{bmatrix} = \begin{bmatrix} 0 & 2 \operatorname{cosec} \theta \\ 2 \operatorname{cosec} \theta & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
This implies $2 \operatorname{cosec} \theta = 0$,which means $\operatorname{cosec} \theta = 0$. Since $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$,it can never be zero for any real $\theta$. Thus,such $\theta_2$ does not exist.

(B) Given $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$.
The characteristic equation is given by $|A - \lambda I| = 0$.
$\begin{vmatrix} 1-\lambda & 0 & 1 \\ 0 & 1-\lambda & 1 \\ 0 & 1 & -\lambda \end{vmatrix} = 0$
$(1-\lambda) [-\lambda(1-\lambda) - 1] = 0$
$(1-\lambda) [-\lambda + \lambda^2 - 1] = 0$
$-\lambda + \lambda^2 - 1 + \lambda^2 - \lambda^3 + \lambda = 0$
$\lambda^3 - 2\lambda^2 + 1 = 0$
By the Cayley-Hamilton theorem,every matrix satisfies its characteristic equation:
$A^3 - 2A^2 + I = 0$
Multiplying both sides by $A^{-1}$:
$A^3 A^{-1} - 2A^2 A^{-1} + I A^{-1} = 0$
$A^2 - 2A + A^{-1} = 0$
$A^{-1} = 2A - A^2$.

(B) Given $A = \frac{1}{7} \begin{bmatrix} 3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3 \end{bmatrix}$.
First,we check if $A$ is an orthogonal matrix by calculating $AA^T$.
$A^T = \frac{1}{7} \begin{bmatrix} 3 & -6 & -2 \\ -2 & -3 & 6 \\ 6 & 2 & 3 \end{bmatrix}$.
$AA^T = \frac{1}{49} \begin{bmatrix} 3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3 \end{bmatrix} \begin{bmatrix} 3 & -6 & -2 \\ -2 & -3 & 6 \\ 6 & 2 & 3 \end{bmatrix}$.
Calculating the product:
Row $1 \times$ Col $1 = (3)(3) + (-2)(-2) + (6)(6) = 9 + 4 + 36 = 49$.
Row $1 \times$ Col $2 = (3)(-6) + (-2)(-3) + (6)(2) = -18 + 6 + 12 = 0$.
Row $1 \times$ Col $3 = (3)(-2) + (-2)(6) + (6)(3) = -6 - 12 + 18 = 0$.
Similarly,all off-diagonal elements result in $0$ and diagonal elements result in $49$.
Thus,$AA^T = \frac{1}{49} (49I) = I$.
Since $AA^T = I$,it follows that $A^{-1} = A^T$.

(C) Given,$A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$.
First,we find the determinant of $A$:
$|A| = (\sin \alpha)(\sin \alpha) - (-\cos \alpha)(\cos \alpha) = \sin^2 \alpha + \cos^2 \alpha = 1$.
Next,we find the adjoint of $A$:
$\text{Adj } A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$.
Since $A^{-1} = \frac{1}{|A|} \text{Adj } A$,we have:
$A^{-1} = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$.
Now,calculate $A + A^{-1}$:
$A + A^{-1} = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} + \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix} = \begin{bmatrix} 2 \sin \alpha & 0 \\ 0 & 2 \sin \alpha \end{bmatrix}$.
Given $A + A^{-1} = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,we equate the corresponding elements:
$2 \sin \alpha = 1 \implies \sin \alpha = \frac{1}{2}$.
Thus,$\alpha = \frac{\pi}{6}$.

(B) Given $A = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}$ and $B^{-1} = \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}$.
First,calculate the product $M = A B^{-1}$:
$M = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} (1)(1) + (2)(0) & (1)(1) + (2)(2) \\ (-2)(1) + (1)(0) & (-2)(1) + (1)(2) \end{bmatrix} = \begin{bmatrix} 1 & 5 \\ -2 & 0 \end{bmatrix}$.
Now,find the inverse $M^{-1} = (A B^{-1})^{-1}$.
The determinant $|M| = (1)(0) - (5)(-2) = 0 + 10 = 10$.
The adjoint $\text{adj}(M) = \begin{bmatrix} 0 & -5 \\ 2 & 1 \end{bmatrix}$.
Thus,$M^{-1} = \frac{1}{|M|} \text{adj}(M) = \frac{1}{10} \begin{bmatrix} 0 & -5 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{1}{2} \\ \frac{1}{5} & \frac{1}{10} \end{bmatrix}$.
Comparing this with $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$,we get $a = 0, b = -\frac{1}{2}, c = \frac{1}{5}, d = \frac{1}{10}$.
Finally,calculate $2b + 5c + 10d$:
$2(-\frac{1}{2}) + 5(\frac{1}{5}) + 10(\frac{1}{10}) = -1 + 1 + 1 = 1$.

(A) Given the characteristic equation $A^3-6A^2+11A-6I=0$.
To find $A^{-1}$,multiply the entire equation by $A^{-1}$:
$A^{-1}(A^3-6A^2+11A-6I) = A^{-1}(0)$
$A^2-6A+11I-6A^{-1} = 0$
Rearranging the terms to solve for $A^{-1}$:
$6A^{-1} = A^2-6A+11I$
$A^{-1} = \frac{1}{6}(A^2-6A+11I)$

(C) Given,$A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$.
First,we calculate the determinant $|A|$:
$|A| = 1(1 - (-3)) - (-1)(2 - (-3)) + 1(2 - 1)$
$|A| = 1(4) + 1(5) + 1(1) = 4 + 5 + 1 = 10$.
Next,we find the cofactor matrix $C$ of $A$:
$C_{11} = (1 - (-3)) = 4, C_{12} = -(2 - (-3)) = -5, C_{13} = (2 - 1) = 1$
$C_{21} = -(-1 - 1) = 2, C_{22} = (1 - 1) = 0, C_{23} = -(1 - (-1)) = -2$
$C_{31} = (3 - 1) = 2, C_{32} = -(-3 - 2) = 5, C_{33} = (1 - (-2)) = 3$.
Thus,$Adj(A) = C^T = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$.
Since $B = A^{-1} = \frac{1}{|A|} Adj(A)$,we have $10 B = Adj(A)$.
Comparing $10 B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$ with $Adj(A) = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$,we get $\alpha = 5$.

(B) Let $A = \begin{bmatrix} 1 & \tan \theta \\ -\tan \theta & 1 \end{bmatrix}$.
Then,$|A| = (1)(1) - (\tan \theta)(-\tan \theta) = 1 + \tan^2 \theta$.
The adjoint of $A$ is $\text{adj}(A) = \begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix}$.
We know that $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1 + \tan^2 \theta} \begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix}$.
Given the equation: $\begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
Substituting $A^{-1}$,we get: $\begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix} \frac{1}{1 + \tan^2 \theta} \begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
Multiplying the matrices: $\frac{1}{1 + \tan^2 \theta} \begin{bmatrix} 1 - \tan^2 \theta & -2 \tan \theta \\ 2 \tan \theta & 1 - \tan^2 \theta \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
This simplifies to: $\begin{bmatrix} \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} & \frac{-2 \tan \theta}{1 + \tan^2 \theta} \\ \frac{2 \tan \theta}{1 + \tan^2 \theta} & \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
Using trigonometric identities $\cos 2 \theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$ and $\sin 2 \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$,we get: $\begin{bmatrix} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
Comparing the elements,we find $a = \cos 2 \theta$ and $b = \sin 2 \theta$.

(D) We know that $\operatorname{det}(kM) = k^n \operatorname{det}(M)$ for an $n \times n$ matrix $M$,and $\operatorname{det}(XY) = \operatorname{det}(X) \operatorname{det}(Y)$.
Given $A = \begin{bmatrix} 2 & 1 & 3 \\ -1 & 2 & 0 \\ 4 & 1 & 3 \end{bmatrix}$.
$|A| = 2(6 - 0) - 1(-3 - 0) + 3(-1 - 8) = 12 + 3 - 27 = -12$.
Given $B = \begin{bmatrix} 3 & 2 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 0 \end{bmatrix}$.
$|B| = 3(0 - 3) - 2(0 - 9) + 1(1 - 6) = -9 + 18 - 5 = 4$.
We need to find $\operatorname{det}(2 B^{-1} A^{-1})$.
Since $A$ and $B$ are $3 \times 3$ matrices,$\operatorname{det}(2 B^{-1} A^{-1}) = 2^3 \operatorname{det}(B^{-1}) \operatorname{det}(A^{-1})$.
Using the property $\operatorname{det}(M^{-1}) = \frac{1}{\operatorname{det}(M)}$,we have:
$\operatorname{det}(2 B^{-1} A^{-1}) = 8 \times \frac{1}{\operatorname{det}(B)} \times \frac{1}{\operatorname{det}(A)}$.
Substituting the values:
$= 8 \times \frac{1}{4} \times \frac{1}{-12} = 2 \times \left(-\frac{1}{12}\right) = -\frac{1}{6}$.

(A) Given the matrix $A = \begin{bmatrix} a+ib & c+id \\ -c+id & a-ib \end{bmatrix}$.
First,we calculate the determinant $|A|$:
$|A| = (a+ib)(a-ib) - (c+id)(-c+id) = (a^2+b^2) - (-(c^2+d^2)) = a^2+b^2+c^2+d^2$.
We know that $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
The adjoint of $A$ is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} a-ib & -c-id \\ c-id & a+ib \end{bmatrix}$.
Thus,$A^{-1} = \frac{1}{a^2+b^2+c^2+d^2} \begin{bmatrix} a-ib & -c-id \\ c-id & a+ib \end{bmatrix}$.
Comparing this with the given $A^{-1} = \begin{bmatrix} a+ib & -c-id \\ -c+id & a-ib \end{bmatrix}$,we observe that for the equality to hold,the determinant $|A|$ must be $1$.
Therefore,$a^2+b^2+c^2+d^2 = 1$.

(D) Given $Q = A^{-1}$.
First,calculate the determinant of $A$:
$|A| = 1(1 - (-3)) - (-1)(2 - (-3)) + 1(2 - 1)$
$|A| = 1(4) + 1(5) + 1(1) = 4 + 5 + 1 = 10$.
Next,find the matrix of cofactors $C_{ij}$:
$C_{11} = +(1+3) = 4, C_{12} = -(2+3) = -5, C_{13} = +(2-1) = 1$
$C_{21} = -(-1-1) = 2, C_{22} = +(1-1) = 0, C_{23} = -(1+1) = -2$
$C_{31} = +(3-1) = 2, C_{32} = -(-3-2) = 5, C_{33} = +(1+2) = 3$
Thus,$\text{Adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$.
Since $A^{-1} = \frac{1}{|A|} \text{Adj } A$,we have:
$Q = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$.
Multiplying by $10$,we get $10Q = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$.
Comparing this with the given $10Q = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & x \\ 1 & -2 & 3 \end{bmatrix}$,we find $x = 5$.
Therefore,option $(d)$ is correct.

(C) Given the diagonal matrix $A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 4 \end{bmatrix}$.
For a diagonal matrix,$A^n = \begin{bmatrix} 3^n & 0 & 0 \\ 0 & 5^n & 0 \\ 0 & 0 & 4^n \end{bmatrix}$.
Therefore,$B = A^3 = \begin{bmatrix} 3^3 & 0 & 0 \\ 0 & 5^3 & 0 \\ 0 & 0 & 4^3 \end{bmatrix} = \begin{bmatrix} 27 & 0 & 0 \\ 0 & 125 & 0 \\ 0 & 0 & 64 \end{bmatrix}$.
The inverse of a diagonal matrix $D = \text{diag}(d_1, d_2, d_3)$ is $D^{-1} = \text{diag}(\frac{1}{d_1}, \frac{1}{d_2}, \frac{1}{d_3})$.
Thus,$B^{-1} = \begin{bmatrix} \frac{1}{27} & 0 & 0 \\ 0 & \frac{1}{125} & 0 \\ 0 & 0 & \frac{1}{64} \end{bmatrix}$.
Hence,option $C$ is correct.

(B) For a square matrix $A$ of order $n$,the following statements are equivalent:
$1$. $A$ is invertible.
$2$. There exists a matrix $B$ such that $AB = I_n$.
$3$. There exists a matrix $C$ such that $CA = I_n$.
If $AB = I_3$,then multiplying by $A^{-1}$ on the right gives $A = I_3 B^{-1}$,which implies $B = A^{-1}$.
Similarly,if $CA = I_3$,then $C = A^{-1}$.
Thus,$B = C = A^{-1}$.
Since all three statements are equivalent,the correct option is $B$.

(B) Given the quadratic equation $x^2-25x+24=0$.
Factoring the equation: $x^2-x-24x+24=0 \Rightarrow x(x-1)-24(x-1)=0 \Rightarrow (x-1)(x-24)=0$.
Thus,the roots are $x=1$ and $x=24$.
Since $A$ is a non-singular matrix,$|A| \neq 0$.
Case $1$: If $k=1$,then $A=\left[\begin{array}{lll}1 & 2 & 1 \\ 3 & 2 & 3 \\ 1 & 1 & 1\end{array}\right]$.
$|A| = 1(2-3) - 2(3-3) + 1(3-2) = -1 - 0 + 1 = 0$.
Since $|A|=0$,$k=1$ is not possible.
Case $2$: If $k=24$,then $A=\left[\begin{array}{ccc}1 & 2 & 1 \\ 3 & 2 & 3 \\ 1 & 1 & 24\end{array}\right]$.
$|A| = 1(48-3) - 2(72-3) + 1(3-2) = 45 - 138 + 1 = -92$.
Now,find the adjoint of $A$:
$C_{11} = 45, C_{12} = -69, C_{13} = 1$
$C_{21} = -47, C_{22} = 23, C_{23} = 1$
$C_{31} = 4, C_{32} = 0, C_{33} = -4$
$\text{adj } A = \left[\begin{array}{ccc}45 & -47 & 4 \\ -69 & 23 & 0 \\ 1 & 1 & -4\end{array}\right]$.
Therefore,$A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{-92} \left[\begin{array}{ccc}45 & -47 & 4 \\ -69 & 23 & 0 \\ 1 & 1 & -4\end{array}\right] = -\frac{1}{92} \left[\begin{array}{ccc}45 & -47 & 4 \\ -69 & 23 & 0 \\ 1 & 1 & -4\end{array}\right]$.

(B) The characteristic equation of a square matrix $A$ is given by $|A-\lambda I|=0$.
$\begin{vmatrix} 1-\lambda & 0 & -2 \\ -2 & -1-\lambda & 2 \\ 3 & 4 & 1-\lambda \end{vmatrix}=0$
Expanding the determinant:
$(1-\lambda)[(-1-\lambda)(1-\lambda)-8] - 2[-8 - 3(-1-\lambda)] = 0$
$(1-\lambda)[-(1-\lambda^2)-8] - 2[-8 + 3 + 3\lambda] = 0$
$(1-\lambda)(\lambda^2-9) - 2(3\lambda-5) = 0$
$\lambda^2 - 9 - \lambda^3 + 9\lambda - 6\lambda + 10 = 0$
$-\lambda^3 + \lambda^2 + 3\lambda + 1 = 0 \Rightarrow \lambda^3 - \lambda^2 - 3\lambda - 1 = 0$
By the Cayley-Hamilton theorem,every square matrix satisfies its characteristic equation:
$A^3 - A^2 - 3A - I = 0$
Multiplying by $A^{-1}$ on both sides:
$A^{-1}(A^3 - A^2 - 3A - I) = 0$
$A^2 - A - 3I - A^{-1} = 0$
$A^{-1} = A^2 - A - 3I$