If $A = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$,then the value of $\alpha$ for which $A^2 = B$ is:

The number of matrices of order $3 \times 3$,whose entries are either $0$ or $1$ and the sum of all the entries is a prime number,is:

Let $A = \begin{bmatrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 & n \\ 0 & n & 0 \\ n & 0 & 0 \end{bmatrix}$. Then,$A^2 + B^2 + AB =$

If $A$ is a nilpotent matrix of index $2$,then $A(I_2+A)^{51}$ is equal to (where $I_2$ is the identity matrix of order $2$)

If $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then for all $n \in N$, find $A^n$.

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?