If $A = \begin{bmatrix} 1 & 4 & 4 \\ 4 & 1 & 4 \\ 4 & 4 & 1 \end{bmatrix}$,then $A^2 - 6A =$ . . . . . . (in $I_3$)

Let $A$ and $B$ be two symmetric matrices of the same order. Then,the matrix $AB - BA$ is

If $A = \begin{bmatrix} 1 & 2 & x \\ 3 & -1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} y \\ x \\ 1 \end{bmatrix}$ are such that $AB = \begin{bmatrix} 6 \\ 8 \end{bmatrix}$,then:

If $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$,then $(BB^TA)^5$ is equal to

If $ 2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} $,then the values of $ x $ and $ y $ are:

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)\} = $ . . . . . .