If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,find $k$ such that $A^{2} = kA - 2I$.

$\cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{bmatrix} + \sin \theta \begin{bmatrix} \sin \theta & - \cos \theta \\ \cos \theta & \sin \theta \end{bmatrix} = $

If $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -3 \end{bmatrix}$, then $A^2 + B^2=$ . . . . . . .

Let $G(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$. If $x+y=0$,then $G(x) G(y) =$

If $A=\left[\begin{array}{rr}i & -i \\ -i & i\end{array}\right]$ and $B=\left[\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right]$, then find $A^8$. (in $B$)

Compute the indicated products $\left[\begin{array}{ccc}3 & -1 & 3 \\ -1 & 0 & 2\end{array}\right]\left[\begin{array}{cc}2 & -3 \\ 1 & 0 \\ 3 & 1\end{array}\right]$