In a third-order matrix $A$, $a_{ij}$ denotes the element in the $i$-th row and $j$-th column. If $a_{ij} = 0$ for $i = j$, $1$ for $i > j$, and $-1$ for $i < j$, then the matrix is:

Let $n \geq 2$ be an integer. $A = \begin{bmatrix} \cos (2 \pi / n) & \sin (2 \pi / n) & 0 \\ -\sin (2 \pi / n) & \cos (2 \pi / n) & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $I$ is the identity matrix of order $3$. Then,

If $A$ and $B$ are skew-symmetric matrices of the same order, then $AB - BA$ is a . . . . . . .

If $A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 8 & 0 \\ 7 & 1 \end{bmatrix}$ and $A^3 = B$,then $x =$

Let $M$ be a $3 \times 3$ matrix satisfying $M\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$,$M\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}$,and $M\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix}$. Then the sum of the diagonal entries of $M$ is

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