If $A = \left[\begin{array}{cc}1+2 i & i \\ -i & 1-2 i\end{array}\right]$ where $i=\sqrt{-1}$,then $A (\operatorname{adj} A )=\ldots$. (in $I$)

If $A = \begin{bmatrix} 0 & 1 & -1 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{bmatrix}$,then evaluate $(A \cdot (\text{adj } A) \cdot A^{-1}) A$.

If $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right]$ and $\operatorname{adj} A=\left[\begin{array}{ccc}4 & -1 & 1 \\ 8 & -7 & a \\ 9 & -6 & b\end{array}\right]$,then find the values of $a$ and $b$.

The multiplicative inverse of $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ is

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

Let $A = \begin{bmatrix} x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda \end{bmatrix}$,then $A^{-1}$ exists if