Let $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$ and $A^{-1} = \alpha I + \beta A$,where $\alpha, \beta \in \mathbb{R}$ and $I$ is the identity matrix of order $2$. Then $4(\alpha - \beta)$ is:

If $A$ is a non-zero square matrix of order $n$ with $\det(I+A) \neq 0$ and $A^3=O$, where $I$ and $O$ are the identity and null matrices of order $n \times n$ respectively, then $(I+A)^{-1}$ is equal to

If $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}_{3 \times 3}$, and $A^{-1} = \begin{bmatrix} \gamma & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix}_{3 \times 3}$, then $|\alpha \cdot \beta \cdot \gamma| = $ (where $| \cdot |$ denotes the absolute value)

If $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & a & 3 \\ 3 & 2 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -2 & 0 & b \\ 7 & -1 & -2 \\ c & 1 & 1 \end{bmatrix}$ and if matrix $B$ is the inverse of matrix $A$,then the value of $4a + 2b - c$ is:

The inverse of $\begin{bmatrix} 3 & 5 & 7 \\ 2 & -3 & 1 \\ 1 & 1 & 2 \end{bmatrix}$ is

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