Let $I$ be a unit matrix of order $6$. Let $A = (a_{ij})$ be a square matrix of order $6$ such that $a_{ij} = \begin{cases} 1, & \text{if } i+j=7 \\ 0, & \text{if } i+j \neq 7 \end{cases}$. Then $(A(\text{adj } A) A^{-1}) A^2 = $

Let $A = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix}$,where $i = \sqrt{-1}$. If $M = A^{T}BA$,then the inverse of the matrix $AM^{2023}A^{T}$ is $.........$

If matrix $A$ and its inverse $A^{-1}$ are given by $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & y \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{bmatrix}$, then the polar coordinates of the point whose Cartesian coordinates are $(x, y)$ are ...

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

If $A = \begin{bmatrix} 8 & -2 \\ -4 & 1 \end{bmatrix}$ is given,then $A^{-1}$ is:

If $A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}$,then $A^T \cdot A^{-1} = $