$A = [a_{ij}]$ is a $3 \times 3$ matrix with positive integers as its elements. Elements of $A$ are such that the sum of all the elements of each row is equal to $6$ and $a_{22} = 2$. If $a_{ii} = \begin{cases} a_{ij} + a_{ji}, & j = i + 1 \text{ when } i < 3 \\ a_{ij} + a_{ji}, & j = 4 - i \text{ when } i = 3 \end{cases}$ for $i = 1, 2, 3$,then $|A| = $

If $A = \begin{bmatrix} 1 & a & 3 \\ b & 2 & c \\ 3 & d & 4 \end{bmatrix}$ is a symmetric matrix and $B = \begin{bmatrix} 0 & 5 & b \\ -5 & 0 & -7 \\ 6 & c & 0 \end{bmatrix}$ is a skew-symmetric matrix,then $AB = $

Let $J_{n, m}=\int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m}-1} d x, \quad \forall n>m$ and $n, m \in N$. Consider a matrix $A=\left[a_{i j}\right]_{3 \times 3}$ where $a_{i j}=J_{6+i, 3}-J_{i+3,3}$ for $i \leq j$ and $a_{i j}=0$ for $i>j$. Then $\left|\operatorname{adj} A^{-1}\right|$ is:

Let $A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{bmatrix}$. The sum of the prime factors of $|P^{-1}AP - 2I|$ is equal to

Let $M = \begin{bmatrix} 0 & 1 & a \\ 1 & 2 & 3 \\ 3 & b & 1 \end{bmatrix}$ and $\operatorname{adj} M = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}$ where $a$ and $b$ are real numbers. Which of the following options is/are correct?
$(1)$ $a+b=3$
$(2)$ $\operatorname{det}(\operatorname{adj} M^2) = 81$
$(3)$ $(\operatorname{adj} M)^{-1} + \operatorname{adj} M^{-1} = -M$
$(4)$ If $M \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$,then $\alpha - \beta + \gamma = 3$

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix}$ and $B = [b_{ij}], 1 \leq i, j \leq 3$. If $B = A^{99} - I$,then the value of $\frac{b_{31} - b_{21}}{b_{32}}$ is: