The inverse of matrix $\begin{bmatrix} 1+pq & p & 0 \\ q & 1+pq & p \\ 0 & q & 1 \end{bmatrix}$ is ...

If matrix A and its inverse $A^{-1}$ are given by $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} \dfrac{1}{2} & -\dfrac{1}{2} & \dfrac{1}{2} \\ -4 & 3 & y \\ \dfrac{5}{2} & -\dfrac{3}{2} & \dfrac{1}{2} \end{bmatrix}$, then the polar co-ordinates of the points whose Cartesian co-ordinates are $(x, y)$ are $\ldots$

Let $A = \begin{bmatrix} -3 & 2 \\ 1 & 4 \end{bmatrix}$ and if $A^2 - 2A + I = \begin{bmatrix} 18 & p \\ q & 11 \end{bmatrix}$, then $\ldots$

Let $A = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$. If $B = \text{adj}\,A$, then the matrix $B^{-1}$ is equal to...