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

If $A = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(x\pi) & \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) \end{bmatrix}$ and $B = \begin{bmatrix} -\frac{1}{\pi} \cos^{-1}(x\pi) & \frac{1}{\pi} \tan^{-1}(\frac{x}{\pi}) \\ \frac{1}{\pi} \sin^{-1}(\frac{x}{\pi}) & -\frac{1}{\pi} \tan^{-1}(\pi x) \end{bmatrix}$,then $A-B$ is equal to:

Construct a $3 \times 2$ matrix whose elements are given by $a_{ij} = \frac{1}{2}|i - 3j|$.

If $\left[\begin{array}{cc}x-1 & 2y \\ x+y & 3\end{array}\right]=\left[\begin{array}{cc}3x-7 & y^2-3 \\ 6 & y\end{array}\right]$,then $\{(x, y)\} = $ . . . . . .

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

Let $A, B, C, D$ and $E$ be $n \times n$ matrices,each with non-zero determinant. If $ABCDE=I$,then $C^{-1}=$