If $A = \begin{vmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{vmatrix}$,then $1 + A^2 =$ . . . . . . .

Which of the given values of $x$ and $y$ make the following pair of matrices equal?
$\left[\begin{array}{cc}3x+7 & 5 \\ y+1 & 2-3x\end{array}\right]=\left[\begin{array}{cc}0 & y-2 \\ 8 & 4\end{array}\right]$

For matrices $A$ and $B$,$A^{\prime} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ and $B^{\prime} = \begin{bmatrix} 4 & 3 & 2 \end{bmatrix}$,then $(BA)^{\prime}$ is . . . . . . .

The number of symmetric matrices of order $3 \times 3$,with all the entries from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$,is:

For the matrices $A$ and $B$,verify that $(AB)^{\prime} = B^{\prime}A^{\prime}$ where $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$.

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?