CUET Mathematics 2022 23 Aug Shift 1 PYQs

If $A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}$, then the matrix AB is equal to

Let $\begin{vmatrix} 3 & y \\ x & 1 \end{vmatrix} = \begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix}$ and $x, y$ are natural numbers, then the number of solutions for the system is:

$\begin{vmatrix} a+b & 1 & 0 \\ a^2-b^2 & a-b & 1 \\ a^3+b^3 & a^2+b^2+ab & a^2-b^2 \end{vmatrix} =$