If $A_1, A_3, \dots, A_{2n-1}$ are $n$ skew-symmetric matrices of the same order,then $B = \sum_{r=1}^n (2r-1)(A_{2r-1})^{2r-1}$ will be:

If $A = \begin{bmatrix} 3 & 4 \\ 1 & -6 \end{bmatrix}$ and $B = \begin{bmatrix} -2 & 5 \\ 6 & 1 \end{bmatrix}$,then find $X$ such that $A + 2X = B$.

Show that $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \ne \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$

Find $AB,$ if $A=\left[\begin{array}{ll}6 & 9 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{lll}2 & 6 & 0 \\ 7 & 9 & 8\end{array}\right].$

The number of all possible matrices of order $3 \times 3$ with each entry $0$ or $1$ is:

If $A = \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 1 \end{bmatrix}$,then $(AB)^T = $