If $\operatorname{adj}\begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b \end{bmatrix}$, then $[a \quad b]$ is equal to

If the adjoint of a $3 \times 3$ matrix $P$ is $\begin{bmatrix} 1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3 \end{bmatrix}$,then the possible value$(s)$ of the determinant of $P$ is (are):

If $B = \begin{bmatrix} 3 & \alpha & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$ is the adjoint of a $3 \times 3$ matrix $A$ and $|A| = 4$,then $\alpha$ is equal to:

For matrix $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$,$(A^{-1})^2 = $ . . . . . .

If $A = [a_{ij}]_{3 \times 3}$, where $a_{ij} = \begin{cases} 1, \text{ if } i + j \text{ is even} \\ 0, \text{ if } i + j \text{ is odd} \end{cases}$, then $adj(A) = \dots$

Let $P = \begin{bmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{bmatrix}$ where $\alpha \in R$. Suppose $Q = [q_{ij}]$ is a matrix satisfying $PQ = kI_3$ for some non-zero $k \in R$. If $q_{23} = -\frac{k}{8}$ and $|Q| = \frac{k^2}{2}$,then $\alpha^2 + k^2$ is equal to?