If $A$ is a non-singular matrix and $(A+I)(A-I)=0$,then $A+A^{-1} = \dots$

Inverse of the matrix $\left[\begin{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right]$ is

If $A$ is an invertible matrix of order $n$, then the determinant of $\operatorname{adj} A$ is equal to :

Let $A$ be a $3 \times 3$ matrix such that $A \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$. Then $A^{-1}$ is

Suppose $A$ is any $3 \times 3$ non-singular matrix and $(A-3 I)(A-5 I)=O$,where $I=I_3$ and $O=O_3$. Here $O_3$ represents the zero matrix of order $3$ and $I_3$ is the identity matrix of order $3$. If $\alpha A+\beta A^{-1}=4 I$,then $\alpha+\beta$ is equal to:

If $A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$,then the inverse of the matrix $A^3$ is