If $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$ and $A \text{ adj } A = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$,then $k$ is equal to :-

Let $A$ be an invertible square matrix of order $3 \times 3$. Then $|(\text{adj} A) \cdot A|$ is

Let $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & -13 \\ 5 & -10 \end{bmatrix}$ be two matrices. If the variables $x$ and $y$ satisfy the matrix equation $((A^{-1})^2 + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$, then the ordered pair $(x, y) =$?

If $A$ is a $n \times n$ matrix,then $adj(adj \,A) = $

If $a, b, c$ and $d$ are real numbers such that $a^2+b^2+c^2+d^2=1$ and $A=\left[\begin{array}{cc}a+ib & c+id \\ -c+id & a-ib\end{array}\right]$, then $A^{-1}$ is equal to

Find the inverse of the matrix (if it exists): $\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$