If $\mathrm{A}=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right],$ prove that $\mathrm{A}^{n}=\left[\begin{array}{lll}3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1}\end{array}\right], n \in \mathrm{N}$.

Let $A = \left[ {\begin{array}{*{20}{c}}
1&2&3\\
2&2&{ – 1}\\
3&0&k
\end{array}} \right]$ and $f(x) = {x^3} – 2{x^2} – \alpha x + \beta  = 0$ . If $A$ satisfies $f(x)=0$ ,then

$A = \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right]$ and $I = \left[ {\begin{array}{*{20}{c}}
1&0\\0&1\end{array}} \right]$ , then which of the following holds for all $n \geq 2, n \in N$ ?

If $A = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\2&{ – 1}\end{array}} \right],\,\,B = \left[ {\begin{array}{*{20}{c}}a&1\\b&{ – 1}\end{array}} \right]$ and ${(A + B)^2} = {A^2} + {B^2}$, then the value of $a$ and $b$ are

Let $M$ be a $3 \times 3$ matrix satisfying $M\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right], \quad M\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{c}1 \\ 1 \\ -1\end{array}\right]$, and $M\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{c}0 \\ 0 \\ 12\end{array}\right]$ Then the sum of the diagonal entries of $M$ is