Adjoint and inverse of matrices Questions in English

(C) We know that for any invertible matrix $M$, $\operatorname{adj}(M) = |M| M^{-1}$.
Let $M = Q^{-1} B P^{-1}$.
Then $\operatorname{adj}(M) = |Q^{-1} B P^{-1}| (Q^{-1} B P^{-1})^{-1}$.
Using the properties of determinants $|XY| = |X||Y|$ and $|X^{-1}| = \frac{1}{|X|}$, we have $|Q^{-1} B P^{-1}| = |Q^{-1}| |B| |P^{-1}| = \frac{1}{|Q|} |B| \frac{1}{|P|}$.
Since $|P| = 1$ and $|Q| = 1$, we get $|Q^{-1} B P^{-1}| = |B|$.
Now, calculating the inverse: $(Q^{-1} B P^{-1})^{-1} = (P^{-1})^{-1} B^{-1} (Q^{-1})^{-1} = P B^{-1} Q$.
Substituting these into the adjoint formula:
$\operatorname{adj}(Q^{-1} B P^{-1}) = |B| P B^{-1} Q$.
Since $\operatorname{adj} B = |B| B^{-1} = A$, we substitute $A$ for $|B| B^{-1}$.
Therefore, $\operatorname{adj}(Q^{-1} B P^{-1}) = P A Q$.

(B) We know that for a $3 \times 3$ matrix $A$, the determinant of its adjoint matrix is given by $\det(\text{adj}(A)) = (\det(A))^{n-1}$, where $n$ is the order of the matrix.
Here, $n = 3$, so $\det(P) = (\det(A))^{3-1} = (\det(A))^2$.
Given $\det(A) = 4$, we have $\det(P) = 4^2 = 16$.
Now, calculate the determinant of matrix $P = \begin{bmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{bmatrix}$:
$\det(P) = 1(3 \times 4 - 3 \times 4) - \alpha(1 \times 4 - 3 \times 2) + 3(1 \times 4 - 3 \times 2)$
$\det(P) = 1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$
$\det(P) = 0 - \alpha(-2) + 3(-2) = 2\alpha - 6$.
Equating the two values: $2\alpha - 6 = 16$.
$2\alpha = 22 \Rightarrow \alpha = 11$.

(C) We know that for any square matrix $M$, the adjoint of the transpose is equal to the transpose of the adjoint.
That is, $\text{adj}(M^{\prime}) = (\text{adj } M)^{\prime}$.
Therefore, $\text{adj}(M^{\prime}) - (\text{adj } M)^{\prime} = (\text{adj } M)^{\prime} - (\text{adj } M)^{\prime} = O$, where $O$ is the null matrix.

(C) We know that for a square matrix $A$ of order $n$, the determinant of its adjoint matrix is given by $|\operatorname{adj} A| = |A|^{n-1}$.
Given that $A$ is a $3 \times 3$ matrix, so $n = 3$.
Therefore, $|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
Given $|B| = |\operatorname{adj} A| = 64$.
So, $|A|^2 = 64$.
Taking the square root on both sides, we get $|A| = \pm \sqrt{64} = \pm 8$.

(B) Let $X$ be a column vector of order $5 \times 1$ where all elements are $1$, i.e.,$X = [1, 1, 1, 1, 1]^T$.
Given that the sum of the elements of each row of $P$ is $1$, we can write this as $PX = X$.
Since $P$ is a non-singular matrix, $P^{-1}$ exists.
Multiplying both sides by $P^{-1}$, we get $P^{-1}(PX) = P^{-1}X$.
This simplifies to $(P^{-1}P)X = P^{-1}X$, which is $IX = P^{-1}X$.
Thus, $P^{-1}X = X$.
This implies that the sum of the elements of each row of $P^{-1}$ is also $1$.

(C) Given $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ and $I_{3} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Then $A-3I_{3} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{bmatrix}$.
To check if the matrix is invertible, we calculate its determinant:
$\det(A-3I_{3}) = -2((-2)(-2) - (1)(1)) - 1((1)(-2) - (1)(1)) + 1((1)(1) - (-2)(1))$
$\det(A-3I_{3}) = -2(4-1) - 1(-2-1) + 1(1+2)$
$\det(A-3I_{3}) = -2(3) - 1(-3) + 1(3) = -6 + 3 + 3 = 0$.
Since the determinant of the matrix is $0$, the matrix $A-3I_{3}$ is non-invertible.

(A) Given matrix $A = \begin{bmatrix} 1 & 2 \\ -4 & -1 \end{bmatrix}$.
First, we calculate the determinant of $A$:
$|A| = (1)(-1) - (2)(-4) = -1 + 8 = 7$.
Since $|A| \neq 0$, $A^{-1}$ exists.
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, $\text{adj}(A) = \begin{bmatrix} -1 & -2 \\ 4 & 1 \end{bmatrix}$.
The inverse is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
$A^{-1} = \frac{1}{7} \begin{bmatrix} -1 & -2 \\ 4 & 1 \end{bmatrix}$.

(B) Given the equation $AB = 3I$, where $A$ and $B$ are square matrices of the same order and $I$ is the identity matrix.
To find $A^{-1}$, we multiply both sides of the equation by $A^{-1}$ from the left:
$A^{-1}(AB) = A^{-1}(3I)$
Using the associative property of matrix multiplication, we get:
$(A^{-1}A)B = 3(A^{-1}I)$
Since $A^{-1}A = I$ and $A^{-1}I = A^{-1}$, the equation simplifies to:
$IB = 3A^{-1}$
$B = 3A^{-1}$
Dividing both sides by $3$, we obtain:
$A^{-1} = \frac{1}{3}B$

(B) Given the equation $A^2-A+I=0$.
We can rewrite this as $A^2-A = -I$.
Multiplying both sides by $A^{-1}$ from the right, we get $(A^2-A)A^{-1} = -I \cdot A^{-1}$.
This simplifies to $A^2 A^{-1} - A A^{-1} = -A^{-1}$.
Since $A A^{-1} = I$, we have $A I - I = -A^{-1}$.
This gives $A - I = -A^{-1}$.
Multiplying by $-1$ on both sides, we get $A^{-1} = I - A$.

(B) Given $q_{ij} = 2^{(i+j-1)}p_{ij}$. The matrix $Q$ can be written as:
$Q = \begin{bmatrix} 2^1 p_{11} & 2^2 p_{12} & 2^3 p_{13} \\ 2^2 p_{21} & 2^3 p_{22} & 2^4 p_{23} \\ 2^3 p_{31} & 2^4 p_{32} & 2^5 p_{33} \end{bmatrix}$
Taking common factors from each row:
$\det(Q) = (2^1 \cdot 2^2 \cdot 2^3) \begin{vmatrix} p_{11} & p_{12} & p_{13} \\ 2 p_{21} & 2 p_{22} & 2 p_{23} \\ 2^2 p_{31} & 2^2 p_{32} & 2^2 p_{33} \end{vmatrix} = 2^6 \cdot (1 \cdot 2 \cdot 2^2) \det(P) = 2^6 \cdot 2^3 \det(P) = 2^9 \det(P)$
Given $\det(Q) = 2^{10}$, so $2^9 \det(P) = 2^{10} \implies \det(P) = 2$.
We know that $\det(\text{adj}(\text{adj } P)) = \det(P)^{(n-1)^2}$, where $n=3$.
$\det(\text{adj}(\text{adj } P)) = \det(P)^{(3-1)^2} = \det(P)^4 = 2^4 = 16$.

(D) First, simplify the integrand: $f(x) = \int \frac{x^{18}(7x^{-8} + 9x^{-10})}{(x^9(x^{-9} + x^{-7} + 2))^2} dx = \int \frac{7x^{-8} + 9x^{-10}}{(x^{-9} + x^{-7} + 2)^2} dx$.
Let $t = x^{-9} + x^{-7} + 2$, then $dt = (-9x^{-10} - 7x^{-8}) dx$, so $-(7x^{-8} + 9x^{-10}) dx = dt$.
Thus, $f(x) = \int -t^{-2} dt = t^{-1} + C = \frac{1}{x^{-9} + x^{-7} + 2} + C = \frac{x^9}{1 + x^2 + 2x^9} + C$.
Given $\lim_{x \to 0} f(x) = 0$, we find $C = 0$. Also $f(1) = \frac{1}{1+1+2} = \frac{1}{4}$, which is consistent.
Now, $f'(x) = \frac{9x^8(1+x^2+2x^9) - x^9(2x + 18x^8)}{(1+x^2+2x^9)^2}$.
At $x=1$, $f'(1) = \frac{9(4) - 1(20)}{4^2} = \frac{36-20}{16} = 1$.
Matrix $A = \begin{bmatrix} 0 & 0 & 1 \\ 1/4 & 1 & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$.
$|A| = 1(\frac{1}{4} \times 4 - \alpha^2 \times 1) = 1 - \alpha^2$.
Given $|B| = |\text{adj}(\text{adj } A)| = |A|^{(n-1)^2} = |A|^4 = 81$, so $|A| = \pm 3$.
$1 - \alpha^2 = 3 \Rightarrow \alpha^2 = -2$ (not possible for real $\alpha^2$) or $1 - \alpha^2 = -3 \Rightarrow \alpha^2 = 4$.

(B) Given $A+A^T=O$, $A$ is a skew-symmetric matrix. Let $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$.
From $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, we get:
$-a = 3 \Rightarrow a = -3$
$-b+c = 2$
$3a + 2b = -3 \Rightarrow 3(-3) + 2b = -3 \Rightarrow 2b = 6 \Rightarrow b = 3$.
Then $c = 2+b = 5$.
So, $A = \begin{bmatrix} 0 & -3 & 3 \\ 3 & 0 & 5 \\ -3 & -5 & 0 \end{bmatrix}$.
Then $A+I = \begin{bmatrix} 1 & -3 & 3 \\ 3 & 1 & 5 \\ -3 & -5 & 1 \end{bmatrix}$.
$|A+I| = 1(1+25) + 3(3+15) + 3(-15+3) = 26 + 54 - 36 = 44$.
We need $\det(\text{adj}(2\text{adj}(A+I)))$.
Since $A+I$ is a $3 \times 3$ matrix, $\text{adj}(A+I)$ is also $3 \times 3$.
$\det(2\text{adj}(A+I)) = 2^3 |\text{adj}(A+I)| = 8 |A+I|^2 = 8(44)^2$.
Then $\det(\text{adj}(2\text{adj}(A+I))) = (8 \cdot 44^2)^2 = (2^3 \cdot (2^2 \cdot 11)^2)^2 = (2^3 \cdot 2^4 \cdot 11^2)^2 = (2^7 \cdot 11^2)^2 = 2^{14} \cdot 11^4$.
Comparing with $(2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, we have $\alpha=14, \beta=0, \gamma=4$.
Thus, $\alpha+\beta+\gamma = 14+0+4 = 18$.

(C) We know that the fundamental property of the adjoint of a matrix is $(\text{adj} A) \cdot A = |A|I$, where $I$ is the identity matrix of order $3 \times 3$.
Taking the determinant on both sides, we get $|(\text{adj} A) \cdot A| = ||A|I|$.
Since $|A|$ is a scalar, we use the property $|kA| = k^n|A|$, where $n$ is the order of the matrix.
Here, $n = 3$, so $|(\text{adj} A) \cdot A| = |A|^3 |I|$.
Since the determinant of an identity matrix $|I| = 1$, we have $|(\text{adj} A) \cdot A| = |A|^3 \times 1 = |A|^3$.

(B) The inverse of a matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Given $A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}$, the determinant $|A| = (2)(-4) - (3)(1) = -8 - 3 = -11$.
Thus, $A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4/11 & 3/11 \\ 1/11 & -2/11 \end{bmatrix}$.
Comparing this with the given $A^{-1} = \begin{bmatrix} a & 3/11 \\ 1/11 & b \end{bmatrix}$, we get $a = 4/11$ and $b = -2/11$.
Therefore, $a+b = 4/11 + (-2/11) = 2/11$.

(D) First, calculate the determinant of $A$: $\det(A) = 1(0-3) - 1(-10-1) + 2(-6-0) = -3 + 11 - 12 = -4$.
Using the property $\text{adj}(\text{adj}(M)) = \det(M)^{n-2} M$ for an $n \times n$ matrix, here $n=3$, so $\text{adj}(\text{adj}(M)) = \det(M) M$.
Let $M = 2(\text{adj} A)^{-1}$. Since $\text{adj} A = \det(A) A^{-1}$, we have $M = 2(\det(A) A^{-1})^{-1} = 2 \det(A)^{-1} A = 2(-4)^{-1} A = -\frac{1}{2} A$.
Then $\text{adj}(\text{adj}(M)) = \det(M) M = \det(-\frac{1}{2} A) (-\frac{1}{2} A) = (-\frac{1}{2})^3 \det(A) (-\frac{1}{2} A) = \frac{1}{16} \det(A) A = \frac{1}{16} (-4) A = -\frac{1}{4} A$.
The sum of all elements of $A$ is $1+1+2-2+0+1+1+3+5 = 12$.
Therefore, the sum of all elements of $-\frac{1}{4} A$ is $-\frac{1}{4} \times 12 = -3$.

(A) Given $A = \begin{bmatrix} 2 & -2 \\ 4 & -2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix}$.
First, calculate the determinant of $A$: $|A| = (2)(-2) - (-2)(4) = -4 + 8 = 4$.
Since $|A| \neq 0$, $A^{-1}$ exists. $A^{-1} = \frac{1}{4} \begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -0.5 & 0.5 \\ -1 & 0.5 \end{bmatrix}$.
Given $PA = B \implies P = BA^{-1} = \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} -0.5 & 0.5 \\ -1 & 0.5 \end{bmatrix} = \begin{bmatrix} -1.5-9 & 1.5+4.5 \\ -0.5-3 & 0.5+1.5 \end{bmatrix} = \begin{bmatrix} -10.5 & 6 \\ -3.5 & 2 \end{bmatrix}$.
Given $AQ = B \implies Q = A^{-1}B = \begin{bmatrix} -0.5 & 0.5 \\ -1 & 0.5 \end{bmatrix} \begin{bmatrix} 3 & 9 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} -1.5+0.5 & -4.5+1.5 \\ -3+0.5 & -9+1.5 \end{bmatrix} = \begin{bmatrix} -1 & -3 \\ -2.5 & -7.5 \end{bmatrix}$.
Now, $P+Q = \begin{bmatrix} -10.5-1 & 6-3 \\ -3.5-2.5 & 2-7.5 \end{bmatrix} = \begin{bmatrix} -11.5 & 3 \\ -6 & -5.5 \end{bmatrix}$.
Then $2(P+Q) = \begin{bmatrix} -23 & 6 \\ -12 & -11 \end{bmatrix}$.
The sum of the diagonal elements is $-23 + (-11) = -34$. The absolute value is $|-34| = 34$.

(B) First, calculate the determinant of $A$: $|A| = -1(0-0) - 1(1-0) - 1(0-0) = -1$.
Since $A$ is a $3 \times 3$ matrix, the property $adj(adj(A)) = |A|^{n-2} A$ holds, where $n=3$. Thus, $adj(adj(A)) = |A|^{3-2} A = (-1)A = -A$.
Next, we know $adj(A) = |A|A^{-1}$. Therefore, $adj(A)(adj(adj(A))) = (|A|A^{-1})(|A|A) = |A|^2 I = (-1)^2 I = I$.
The given equation becomes $A^2 - \alpha A + \beta I = M$, where $M = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$.
Calculate $A^2$: $\begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Substituting these into the equation: $\begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$.
Comparing the elements, for the element at $(1, 2)$: $-1 - \alpha = -2 \implies \alpha = 1$.
For the element at $(1, 3)$: $1 + \alpha = 2 \implies \alpha = 1$.
For the element at $(3, 3)$: $1 - \alpha + \beta = -1 \implies 1 - 1 + \beta = -1 \implies \beta = -1$.
Thus, $(\alpha - \beta)^2 = (1 - (-1))^2 = 2^2 = 4$.

(A) Given $A = \begin{bmatrix} 2i & i^3 \\ i^2 & 1 \end{bmatrix}$. Since $i^2 = -1$ and $i^3 = -i$, we have $A = \begin{bmatrix} 2i & -i \\ -1 & 1 \end{bmatrix}$.
Step $1$: Find the determinant $|A| = (2i)(1) - (-i)(-1) = 2i - i = i$.
Step $2$: Find the adjoint $adj(A)$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $adj(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Thus, $adj(A) = \begin{bmatrix} 1 & i \\ 1 & 2i \end{bmatrix}$.
Step $3$: Calculate $A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{i} \begin{bmatrix} 1 & i \\ 1 & 2i \end{bmatrix}$.
Since $\frac{1}{i} = -i$, we get $A^{-1} = -i \begin{bmatrix} 1 & i \\ 1 & 2i \end{bmatrix} = \begin{bmatrix} -i & -i^2 \\ -i & -2i^2 \end{bmatrix} = \begin{bmatrix} -i & 1 \\ -i & 2 \end{bmatrix}$.

(B) Given $A = \begin{bmatrix} -1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & -1/2 \end{bmatrix}$.
This matrix is of the form $\begin{bmatrix} \cos(240^\circ) & -\sin(240^\circ) \\ \sin(240^\circ) & \cos(240^\circ) \end{bmatrix}$, which is a rotation matrix $R_\theta$ with $\theta = 240^\circ = 4\pi/3$.
For a rotation matrix, $A^{-1} = R_{-\theta}$ and $A^2 = R_{2\theta}$.
$A^{-1} = \begin{bmatrix} \cos(120^\circ) & -\sin(120^\circ) \\ \sin(120^\circ) & \cos(120^\circ) \end{bmatrix} = \begin{bmatrix} -1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & -1/2 \end{bmatrix}$.
$A^2 = \begin{bmatrix} \cos(480^\circ) & -\sin(480^\circ) \\ \sin(480^\circ) & \cos(480^\circ) \end{bmatrix} = \begin{bmatrix} \cos(120^\circ) & -\sin(120^\circ) \\ \sin(120^\circ) & \cos(120^\circ) \end{bmatrix} = \begin{bmatrix} -1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & -1/2 \end{bmatrix}$.
Thus, $A^{-1} - A^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, which is a null matrix.
$A$ null matrix is also a diagonal matrix and a scalar matrix, but it is not a unit matrix.

(C) Step $1$: Calculate $A^2$.
$A^2 = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Step $2$: Since $A^2 = I$, multiply both sides by $A^{-1}$ to get $A = A^{-1}$.
Step $3$: Check the options. Option $(C)$ states $A^{-1} = A$, which matches our result.

(C) Step $1$: Calculate the determinant of $A$. $|A| = 2(3-0) - 0(15-0) + (-1)(5-0) = 6 - 5 = 1$.
Step $2$: Use the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$. Since $|A| = 1$, $A^{-1} = \text{adj}(A)$.
Step $3$: Find the cofactor matrix $C_{ij}$.
$C_{11} = +(3-0) = 3$, $C_{12} = -(15-0) = -15$, $C_{13} = +(5-0) = 5$.
$C_{21} = -(0 - (-1)) = -1$, $C_{22} = +(6-0) = 6$, $C_{23} = -(2-0) = -2$.
$C_{31} = +(0 - (-1)) = 1$, $C_{32} = -(0 - (-5)) = -5$, $C_{33} = +(2-0) = 2$.
Step $4$: The adjoint is the transpose of the cofactor matrix: $\text{adj}(A) = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$.
Step $5$: Comparing with $A^{-1}$, we get $\gamma = 3$, $\alpha = -15$, $\beta = 5$.
Step $6$: Calculate $|\alpha \cdot \beta \cdot \gamma| = |(-15) \cdot 5 \cdot 3| = |-225| = 225$.

(C) Step $1$: Find the determinant $|A| = 1(0 - (-25)) - 3(0 - (-10)) - 2(-15 - 0) = 1(25) - 3(10) - 2(-15) = 25 - 30 + 30 = 25$.
Step $2$: The element in the $1^{st}$ row and $2^{nd}$ column of $A^{-1}$ is given by $\frac{C_{21}}{|A|}$, where $C_{21}$ is the cofactor of the element at the $2^{nd}$ row and $1^{st}$ column of $A$.
Step $3$: $C_{21} = (-1)^{2+1} \times \text{minor}_{21} = -1 \times \begin{vmatrix} 3 & -2 \\ 5 & 0 \end{vmatrix} = -1 \times (0 - (-10)) = -10$.
Step $4$: The required element is $\frac{C_{21}}{|A|} = \frac{-10}{25} = -\frac{2}{5}$. Note: The provided options do not contain the correct value. Re-evaluating the cofactor: $C_{21} = -10$, $|A|=25$, result is $-2/5$.

(A) Step $1$: Find the determinant of matrix $A$.
$|A| = 1(4 - 9) - 3(12 - 9) + 3(9 - 3) = 1(-5) - 3(3) + 3(6) = -5 - 9 + 18 = 4$.
Step $2$: The element in the third row and second column of $A^{-1}$ is given by $\frac{C_{23}}{|A|}$, where $C_{23}$ is the cofactor of the element in the second row and third column of $A$.
Step $3$: Calculate the minor $M_{23}$ by deleting the second row and third column: $M_{23} = \begin{vmatrix} 1 & 3 \\ 3 & 3 \end{vmatrix} = (3 - 9) = -6$.
Step $4$: The cofactor $C_{23} = (-1)^{2+3} M_{23} = -1 \times (-6) = 6$.
Step $5$: The required element is $\frac{C_{23}}{|A|} = \frac{6}{4} = \frac{3}{2}$.

(A) Step $1$: Find the determinant of $A$. $|A| = (3 \times 4) - (-2 \times 1) = 12 + 2 = 14$.
Step $2$: Find the adjoint of $A$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Step $3$: Thus, $adj(A) = \begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix}$.
Step $4$: The inverse is $A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{14} \begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} \frac{4}{14} & \frac{2}{14} \\ -\frac{1}{14} & \frac{3}{14} \end{bmatrix} = \begin{bmatrix} \frac{2}{7} & \frac{1}{7} \\ -\frac{1}{14} & \frac{3}{14} \end{bmatrix}$.

(C) Given the equation: $A^2 - A + I = 0$
Subtract $I$ from both sides: $A^2 - A = -I$
Factor out $A$ from the left side: $A(A - I) = -I$
Multiply both sides by $A^{-1}$ from the left: $A^{-1}A(A - I) = A^{-1}(-I)$
Since $A^{-1}A = I$, we have: $I(A - I) = -A^{-1}$
$A - I = -A^{-1}$
Multiplying by $-1$ on both sides: $A^{-1} = -(A - I) = I - A$

(D) Step $1$: Find $A^{-1}$. Given $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$, $\det(A) = (3)(-1) - (-4)(1) = -3 + 4 = 1$. Thus, $A^{-1} = \frac{1}{1} \begin{bmatrix} -1 & 4 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 \\ -1 & 3 \end{bmatrix}$.
Step $2$: Calculate $(A^{-1})^2$. $(A^{-1})^2 = \begin{bmatrix} -1 & 4 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} -1 & 4 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} (-1)(-1) + (4)(-1) & (-1)(4) + (4)(3) \\ (-1)(-1) + (3)(-1) & (-1)(4) + (3)(3) \end{bmatrix} = \begin{bmatrix} -3 & 8 \\ -2 & 5 \end{bmatrix}$.
Step $3$: Calculate $(A^{-1})^2 + B$. $\begin{bmatrix} -3 & 8 \\ -2 & 5 \end{bmatrix} + \begin{bmatrix} 6 & -13 \\ 5 & -10 \end{bmatrix} = \begin{bmatrix} 3 & -5 \\ 3 & -5 \end{bmatrix}$.
Step $4$: Solve the system $\begin{bmatrix} 3 & -5 \\ 3 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$. This implies $3x - 5y = 0$, or $3x = 5y$. Among the options, only $(5, 3)$ satisfies $3(5) = 5(3)$, i.e., $15 = 15$.

(C) We know that $A \cdot A^{-1} = I$, where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Multiplying the second row of $A$ by the third column of $A^{-1}$ gives the element at $(2, 3)$ of $I$, which is $0$.
$(1 \times \frac{1}{2}) + (2 \times y) + (3 \times \frac{1}{2}) = 0 \implies \frac{1}{2} + 2y + \frac{3}{2} = 0 \implies 2 + 2y = 0 \implies y = -1$.
Multiplying the third row of $A$ by the first column of $A^{-1}$ gives the element at $(3, 1)$ of $I$, which is $0$.
$(3 \times \frac{1}{2}) + (x \times -4) + (1 \times \frac{5}{2}) = 0 \implies \frac{3}{2} - 4x + \frac{5}{2} = 0 \implies 4 - 4x = 0 \implies x = 1$.
The Cartesian coordinates are $(1, -1)$.
For polar coordinates $(r, \theta)$, $r = \sqrt{x^2 + y^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.
Since the point $(1, -1)$ is in the fourth quadrant, $\theta = \arctan(\frac{y}{x}) = \arctan(-1) = \frac{7\pi}{4}$.
Thus, the polar coordinates are $(\sqrt{2}, \frac{7\pi}{4})$.

(C) Let $A = \begin{bmatrix} 1 + pq & p & 0 \\ q & 1 + pq & p \\ 0 & q & 1 \end{bmatrix}$.
First, calculate the determinant $|A| = (1+pq)((1+pq)(1) - pq) - p(q(1) - 0) + 0 = (1+pq)(1) - pq = 1 + pq - pq = 1$.
Since $|A| = 1$, the inverse $A^{-1}$ is the adjugate matrix $adj(A)$.
The cofactors $C_{ij}$ are:
$C_{11} = (1+pq)(1) - pq = 1$, $C_{12} = -(q - 0) = -q$, $C_{13} = q^2 - 0 = q^2$.
$C_{21} = -(p - 0) = -p$, $C_{22} = (1+pq) - 0 = 1+pq$, $C_{23} = -(q(1+pq) - 0) = -(q+pq^2)$.
$C_{31} = p^2 - 0 = p^2$, $C_{32} = -(p(1+pq) - 0) = -(p+p^2q)$, $C_{33} = (1+pq)^2 - pq = 1 + 2pq + p^2q^2 - pq = 1 + pq + p^2q^2$.
Thus, $A^{-1} = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix} = \begin{bmatrix} 1 & -p & p^2 \\ -q & 1 + pq & -(p + p^2q) \\ q^2 & -(q + pq^2) & 1 + pq + p^2q^2 \end{bmatrix}$.

(D) Step $1$: Calculate the determinant of matrix $A$.
$|A| = 2((-3)(1) - (1)(2)) - (-1)((4)(1) - (1)(1)) + 4((4)(2) - (-3)(1))$
$|A| = 2(-3 - 2) + 1(4 - 1) + 4(8 + 3)$
$|A| = 2(-5) + 1(3) + 4(11) = -10 + 3 + 44 = 37$.
Step $2$: The inverse $A^{-1}$ is given by $\frac{1}{|A|} \text{adj}(A)$.
Step $3$: The element at position $(3,3)$ of the adjoint matrix is the cofactor $C_{33}$.
$C_{33} = (-1)^{3+3} \begin{vmatrix} 2 & -1 \\ 4 & -3 \end{vmatrix} = (1)((2)(-3) - (-1)(4)) = -6 + 4 = -2$.
Step $4$: Since $B = A^{-1} = \frac{1}{37} \text{adj}(A)$, the element at $(3,3)$ in $B$ is $\frac{C_{33}}{37} = \frac{-2}{37}$.
Comparing this with the given matrix $B$, we get $k = -2$.

(A) Step $1$: Find the determinant of $A$. $|A| = (1)(4) - (-5)(-2) = 4 - 10 = -6$.
Step $2$: Find the adjoint of $A$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Step $3$: Thus, $\text{adj}(A) = \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix}$.
Step $4$: Use the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
Step $5$: $A^{-1} = \frac{1}{-6} \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix} = -\frac{1}{6} \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix}$.

(C) Given $A = \begin{bmatrix} \sec \theta & -\tan \theta \\ -\tan \theta & \sec \theta \end{bmatrix}$.
For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\text{adj } A = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Thus, $\text{adj } A = \begin{bmatrix} \sec \theta & \tan \theta \\ \tan \theta & \sec \theta \end{bmatrix}$.
Given $A + \text{adj } A = 4I$, where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
$\begin{bmatrix} \sec \theta & -\tan \theta \\ -\tan \theta & \sec \theta \end{bmatrix} + \begin{bmatrix} \sec \theta & \tan \theta \\ \tan \theta & \sec \theta \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$.
Adding the matrices: $\begin{bmatrix} 2 \sec \theta & 0 \\ 0 & 2 \sec \theta \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$.
Equating the elements: $2 \sec \theta = 4 \implies \sec \theta = 2$.
Since $\sec \theta = 2$, $\cos \theta = \frac{1}{2}$.
For $\theta \in (0, \frac{\pi}{2})$, $\theta = \frac{\pi}{3}$.

(A) Step $1$: Calculate $\det(A)$. Since $A$ is an upper triangular matrix, $\det(A) = (2k - 1)^3$.
Step $2$: Calculate $\det(\text{adj } A)$. We know $\det(\text{adj } A) = (\det A)^{n-1}$. Here $n=3$, so $\det(\text{adj } A) = ((2k - 1)^3)^2 = (2k - 1)^6$.
Step $3$: Calculate $\det(B)$. $B$ is a skew-symmetric matrix of order $3 \times 3$. The determinant of a skew-symmetric matrix of odd order is $0$. Thus, $\det(B) = 0$.
Step $4$: Calculate $\det(\text{adj } B)$. Since $\det(B) = 0$, $\det(\text{adj } B) = (\det B)^{3-1} = 0^2 = 0$.
Step $5$: Solve the equation $(2k - 1)^6 + 0 = 11^6$. This implies $(2k - 1)^6 = 11^6$, so $2k - 1 = 11$ or $2k - 1 = -11$.
Step $6$: If $2k - 1 = 11$, then $2k = 12$, so $k = 6$. Then $k - 5 = 6 - 5 = 1$.
Step $7$: If $2k - 1 = -11$, then $2k = -10$, so $k = -5$. Then $k - 5 = -5 - 5 = -10$ (not in options).
Therefore, the value is $1$.

(C) Given $A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$.
We know that $A(adj A) = |A|I$, where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
$|A| = (5a)(2) - (-b)(3) = 10a + 3b$.
So, $A(adj A) = \begin{bmatrix} 10a + 3b & 0 \\ 0 & 10a + 3b \end{bmatrix}$.
Also, $A^T = \begin{bmatrix} 5a & 3 \\ -b & 2 \end{bmatrix}$.
$AA^T = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 5a & 3 \\ -b & 2 \end{bmatrix} = \begin{bmatrix} 25a^2 + b^2 & 15a - 2b \\ 15a - 2b & 9 + 4 \end{bmatrix} = \begin{bmatrix} 25a^2 + b^2 & 15a - 2b \\ 15a - 2b & 13 \end{bmatrix}$.
Equating $A(adj A) = AA^T$:
$10a + 3b = 13$ (from the $(2,2)$ element).
For the off-diagonal elements, $15a - 2b = 0$, so $b = \frac{15a}{2}$.
Substitute $b$ into $10a + 3b = 13$:
$10a + 3(\frac{15a}{2}) = 13 \implies 10a + \frac{45a}{2} = 13 \implies \frac{20a + 45a}{2} = 13 \implies 65a = 26 \implies a = \frac{26}{65} = \frac{2}{5}$.
Then $b = \frac{15}{2} \times \frac{2}{5} = 3$.
Thus, $5a + b = 5(\frac{2}{5}) + 3 = 2 + 3 = 5$.

(D) We know that for any square matrix $A$ of order $n$, $\text{adj}(\text{adj } A) = |A|^{n-2} A$.
Also, the property of the adjoint matrix is $B = \text{adj } A$, so $B^{-1} = (\text{adj } A)^{-1} = \frac{1}{|A|} A$.
First, calculate the determinant $|A|$:
$|A| = \cos \alpha (\cos \alpha - 0) - (-\sin \alpha)(\sin \alpha - 0) + 0 = \cos^2 \alpha + \sin^2 \alpha = 1$.
Since $|A| = 1$, we have $B = \text{adj } A$.
We know that $A \cdot \text{adj } A = |A| I = I$.
Thus, $\text{adj } A = A^{-1}$.
Therefore, $B = A^{-1}$.
Then $B^{-1} = (A^{-1})^{-1} = A$.

(B) Step $1$: Construct the matrix $A$. For $i, j \in \{1, 2, 3\}$, $a_{ij} = 1$ if $i+j$ is even and $a_{ij} = 0$ if $i+j$ is odd.
$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$.
Step $2$: Calculate the determinant $|A|$.
$|A| = 1(1-0) - 0(0-0) + 1(0-1) = 1 - 1 = 0$.
Step $3$: Since $|A| = 0$, the matrix $A$ is singular. For a singular matrix, the product $A \cdot adj(A) = |A|I = 0$. However, calculating the cofactor matrix $C$:
$C_{11} = 1, C_{12} = 0, C_{13} = -1, C_{21} = 0, C_{22} = 0, C_{23} = 0, C_{31} = -1, C_{32} = 0, C_{33} = 1$.
$adj(A) = C^T = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$. Since this is not among the options, we re-evaluate the question context. Given the structure, the correct answer is the zero matrix if the rank is $1$, but here $adj(A)$ is as calculated.

(C) Step $1$: We know that $adj(A^{-1}) = adj((A^{-1})) = \frac{1}{\det(A)} A$.
Step $2$: Calculate $\det(A)$. Since $A$ is an upper triangular matrix, $\det(A) = (-1) \times (2) \times (-1) = 2$.
Step $3$: Thus, $adj(A^{-1}) = \frac{1}{2} A = \begin{bmatrix} -0.5 & 1012.5 & 1013 \\ 0 & 1 & 1013.5 \\ 0 & 0 & -0.5 \end{bmatrix}$.
Step $4$: Sum of all elements = $-0.5 + 1012.5 + 1013 + 0 + 1 + 1013.5 + 0 + 0 - 0.5 = 3039$.

(A) Given $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
Using the property of rotation matrices, $A^n = \begin{bmatrix} \cos(n\theta) & -\sin(n\theta) \\ \sin(n\theta) & \cos(n\theta) \end{bmatrix}$.
Thus, $A^{-3} = \begin{bmatrix} \cos(-3\theta) & -\sin(-3\theta) \\ \sin(-3\theta) & \cos(-3\theta) \end{bmatrix} = \begin{bmatrix} \cos(3\theta) & \sin(3\theta) \\ -\sin(3\theta) & \cos(3\theta) \end{bmatrix}$.
Given $\theta = \pi/6$, then $3\theta = 3(\pi/6) = \pi/2$.
Substituting the value, $A^{-3} = \begin{bmatrix} \cos(\pi/2) & \sin(\pi/2) \\ -\sin(\pi/2) & \cos(\pi/2) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$.

(B) Given $AB = C$. Multiplying by $A^{-1}$ on the left, we get $B = A^{-1}C$. Taking the inverse of both sides, $B^{-1} = (A^{-1}C)^{-1} = C^{-1}A$.
First, find $C^{-1}$. The determinant $|C| = (7 \times 6) - (3 \times 0) = 42$. The adjoint $adj(C) = \begin{bmatrix} 6 & -3 \\ 0 & 7 \end{bmatrix}$. Thus, $C^{-1} = \frac{1}{42} \begin{bmatrix} 6 & -3 \\ 0 & 7 \end{bmatrix}$.
Now, $B^{-1} = C^{-1}A = \frac{1}{42} \begin{bmatrix} 6 & -3 \\ 0 & 7 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$.
$B^{-1} = \frac{1}{42} \begin{bmatrix} (6 \times 3) + (-3 \times -1) & (6 \times 1) + (-3 \times 2) \\ (0 \times 3) + (7 \times -1) & (0 \times 1) + (7 \times 2) \end{bmatrix} = \frac{1}{42} \begin{bmatrix} 21 & 0 \\ -7 & 14 \end{bmatrix}$.
This does not match the options directly. Re-evaluating: $B = A^{-1}C \implies B^{-1} = C^{-1}A$. Calculation: $C^{-1}A = \frac{1}{42} \begin{bmatrix} 21 & 0 \\ -7 & 14 \end{bmatrix} = \frac{7}{42} \begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix}$.
Thus, the correct option is $B$.

(B) We use the property $(AB)^{-1} = B^{-1}A^{-1}$.
First, find $A^{-1}$. Given $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$, $|A| = (2)(-2) - (3)(5) = -4 - 15 = -19$.
$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{-19} \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix} = \frac{1}{19} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$.
Now, $(AB)^{-1} = B^{-1}A^{-1} = \begin{bmatrix} 1/5 & 2/5 \\ 2/5 & -1/5 \end{bmatrix} \times \frac{1}{19} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$.
$= \frac{1}{5 \times 19} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix} = \frac{1}{95} \begin{bmatrix} (1)(2)+(2)(5) & (1)(3)+(2)(-2) \\ (2)(2)+(-1)(5) & (2)(3)+(-1)(-2) \end{bmatrix}$.
$= \frac{1}{95} \begin{bmatrix} 12 & -1 \\ -1 & 8 \end{bmatrix}$.

(D) Step $1$: Given $XA = B^T$, multiply by $A^{-1}$ on the right: $X = B^T A^{-1}$.
Step $2$: Given $A^{-1}Y = B$, we have $Y = AB$.
Step $3$: Calculate $XY = (B^T A^{-1})(AB) = B^T (A^{-1}A) B = B^T I B = B^T B$.
Step $4$: $B^T = \begin{bmatrix} 1 & 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$.
Step $5$: $XY = \begin{bmatrix} 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = (1 \times 1) + (0 \times 0) + (1 \times 1) = 1 + 0 + 1 = [2]$.

(A) Step $1$: Calculate $2A = 2 \begin{bmatrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & 14 \end{bmatrix}$.
Step $2$: Calculate $3B = 3 \begin{bmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} -3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 12 \end{bmatrix}$.
Step $3$: Calculate $2A + 3B = \begin{bmatrix} 6-3 & 0 & 0 \\ 0 & -10+6 & 0 \\ 0 & 0 & 14+12 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 26 \end{bmatrix}$.
Step $4$: The inverse of a diagonal matrix $D = \text{diag}(d_1, d_2, d_3)$ is $D^{-1} = \text{diag}(1/d_1, 1/d_2, 1/d_3)$.
Step $5$: Thus, $(2A + 3B)^{-1} = \begin{bmatrix} 1/3 & 0 & 0 \\ 0 & -1/4 & 0 \\ 0 & 0 & 1/26 \end{bmatrix}$.

(D) Step $1$: Construct the matrix $A$ using $a_{ij} = |2i - 5j|$ for $i, j \in \{1, 2, 3\}$.
$a_{11} = |2(1) - 5(1)| = |-3| = 3$
$a_{12} = |2(1) - 5(2)| = |-8| = 8$
$a_{13} = |2(1) - 5(3)| = |-13| = 13$
$a_{21} = |2(2) - 5(1)| = |-1| = 1$
$a_{22} = |2(2) - 5(2)| = |-6| = 6$
$a_{23} = |2(2) - 5(3)| = |-11| = 11$
$a_{31} = |2(3) - 5(1)| = |1| = 1$
$a_{32} = |2(3) - 5(2)| = |-4| = 4$
$a_{33} = |2(3) - 5(3)| = |-9| = 9$
So, $A = \begin{pmatrix} 3 & 8 & 13 \\ 1 & 6 & 11 \\ 1 & 4 & 9 \end{pmatrix}$.
Step $2$: Calculate the determinant $|A|$.
$|A| = 3(54 - 44) - 8(9 - 11) + 13(4 - 6) = 3(10) - 8(-2) + 13(-2) = 30 + 16 - 26 = 20$.
Step $3$: The element in the $2^{nd}$ row and $3^{rd}$ column of $A^{-1}$ is $\frac{C_{32}}{|A|}$, where $C_{32}$ is the cofactor of $a_{32}$.
$C_{32} = (-1)^{3+2} \begin{vmatrix} 3 & 13 \\ 1 & 11 \end{vmatrix} = -1(33 - 13) = -20$.
Step $4$: Element $= \frac{-20}{20} = -1$.

(A) Let $t = \tan(\theta/2)$. Then $A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$. The determinant $|A| = 1 + t^2$.
$A^{-1} = \frac{1}{1+t^2} \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$.
Now, $A^{-1}B = \frac{1}{1+t^2} \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} = \frac{1}{1+t^2} \begin{bmatrix} 1-t^2 & 2t \\ -2t & 1-t^2 \end{bmatrix}$.
Using the identities $\cos \theta = \frac{1-t^2}{1+t^2}$ and $\sin \theta = \frac{2t}{1+t^2}$, we get $A^{-1}B = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$.

(C) Step $1$: Statement $I$ is true because for a non-singular matrix, $|A| \neq 0$, which is the necessary and sufficient condition for the existence of the inverse matrix $A^{-1}$.
Step $2$: Statement $II$ is true. Given $A$ and $B$ are symmetric, $A' = A$ and $B' = B$. Consider the transpose of $(AB - BA)$:
$(AB - BA)' = (AB)' - (BA)' = B'A' - A'B' = BA - AB = -(AB - BA)$.
Since the transpose of $(AB - BA)$ is its negative, $(AB - BA)$ is a skew-symmetric matrix.

(A) Step $1$: The fundamental property of the adjoint of a matrix $A$ is $A(\text{adj } A) = (\text{adj } A)A = |A|I$, where $|A|$ is the determinant of $A$ and $I$ is the identity matrix.
Step $2$: Option $(A)$ states $A(\text{adj } A) = (\text{adj } A)A = AI$. This is incorrect because it omits the determinant factor $|A|$.
Step $3$: Option $(B)$ is the correct identity. Option $(C)$ is the standard reversal law for the inverse of a product of matrices. Option $(D)$ is the condition for a matrix to be invertible.
Step $4$: Therefore, the incorrect statement is $(A)$.