Let the matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$ satisfy $A^n = A^{n-2} + A^2 - I$ for $n \geq 3$. Then the sum of all the elements of $A^{50}$ is:

If $A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$,then $AB = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ and $BA = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Clearly $AB \neq BA$. Thus,matrix multiplication is not commutative. Is this statement true for all matrices?

If $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$,then $(aI + bA)^n$ is (where $I$ is the identity matrix of order $2$)

If $A = \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix}$ and $A^2 + xA + yI_2 = O_2$, where $I_2$ and $O_2$ are the identity matrix and null matrix of order $2$ respectively, then:

If $A$ is an $m \times n$ matrix and $B$ is a matrix such that both $AB$ and $BA$ are defined,then the order of $B$ is

If $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}$ is a matrix satisfying the equation $A A^T = 9 I$,where $I$ is the identity matrix,then $a^2 + b^2 =$