If $A=\begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$,then $(\operatorname{Adj} A)^{-1}=$

If $\omega$ is a complex cube root of unity and $A=\begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$,then $A^{-1}=$

Which of the following matrices is invertible?
$A_{1}=\begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix}$
$A_{2}=\begin{bmatrix} -1 & -2 & 3 \\ 4 & 5 & 7 \\ 2 & 4 & -6 \end{bmatrix}$
$A_{3}=\begin{bmatrix} 1 & 0 & 0 \\ 5 & 2 & 1 \\ 7 & 2 & 1 \end{bmatrix}$
$A_{4}=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix}$

Let $A$ be a $4 \times 4$ matrix and $P$ be its adjoint matrix. If $|P|=\left|\frac{A}{2}\right|$,then $\left|A^{-1}\right|=$

The inverse of the matrix $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is

If $A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$,then verify that $A \text{ adj } A = |A| I$. Also,find $A^{-1}$.