If $A = [a\, b]$,$B = [-b\, -a]$ and $C = \begin{bmatrix} a \\ -a \end{bmatrix}$,then the correct statement is

If $A=\left[\begin{array}{lll}9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2\end{array}\right]$ and $AA^T-A^2=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$, then $\sum_{\substack{1 \leq i \leq 3 \\ 1 \leq j \leq 3}} a_{i j}=$

Let $A = [a_{ij}]$ be a real matrix of order $3 \times 3$,such that $a_{i1} + a_{i2} + a_{i3} = 1$,for $i = 1, 2, 3$. Then,the sum of all the entries of the matrix $A^3$ is equal to:

Let $A, B, C, D$ and $E$ be $n \times n$ matrices,each with non-zero determinant. If $ABCDE=I$,then $C^{-1}=$

The order of the matrix product $\left[ {x\,y\,z} \right]\,\left[ {\begin{array}{*{20}{c}} a&h&g\\ h&b&f\\ g&f&c \end{array}} \right]\,\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right]$ is:

If $A = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix}$,$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $A^{2} = 8A + kI$,then the value of $k$ is