If $A = \begin{bmatrix} k & 2 \\ -2 & -k \end{bmatrix}$,then $A^{-1}$ does not exist if $k =$

If $A(\alpha) = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$,then $[A^2(\alpha)]^{-1} = $

Let $A = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ satisfy $A^2 + \alpha(adj(adj(A))) + \beta(adj(A)(adj(adj(A)))) = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$ for some $\alpha, \beta \in R$. Then $(\alpha - \beta)^2$ is equal to . . . . . . .

If $A = \begin{bmatrix} -2 & 6 \\ -5 & 7 \end{bmatrix}$,then find $adj(A)$.

Obtain the inverse of the following matrix using elementary row operations: $A = \left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$

If $\operatorname{adj}\begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b \end{bmatrix}$, then $[a \quad b]$ is equal to