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

Let $A$ be a $2 \times 2$ symmetric matrix such that $A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$ and the determinant of $A$ is $1$. If $A^{-1} = \alpha A + \beta I$,where $I$ is an identity matrix of order $2 \times 2$,then $\alpha + \beta$ equals:

If $A=\begin{bmatrix} 2a & -3b \\ 3 & 2 \end{bmatrix}$ and $A \cdot \operatorname{adj} A = A A^{T}$,then $2a + 3b$ is

The inverse matrix of $\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$ is

If $(BA)^{-1} = C$ where $B = \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1 \end{bmatrix}$ and $C = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix}$,then $A^{-1}$ is given by

If $A$ is a matrix of order $2$ and $I$ is the identity matrix of order $2$ such that $A^2 - 4A + 3I = 0$,then $(A + 3I)^{-1} =$