The number of elements in the set $\{A=\begin{bmatrix} a & b \\ 0 & d \end{bmatrix} : a, b, d \in \{-1, 0, 1\} \text{ and } (I-A)^3 = I-A^3 \}$,where $I$ is the $2 \times 2$ identity matrix,is:

Let $M$ and $N$ be two invertible square matrices over $\mathbb{R}$ of order $2$ such that $N$ is diagonal. Then $M N M^{-1}$ is diagonal . . . . . .

If $A=\left[\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$,then $A+A^3+A^4+A^5+3 I=$

If ${\Delta _r} = \left| {\begin{array}{*{20}{c}} r&{2r - 1}&{3r - 2} \\ {\frac{n}{2}}&{n - 1}&a \\ {\frac{1}{2}n\left( {n - 1} \right)}&{{{\left( {n - 1} \right)}^2}}&{\frac{1}{2}\left( {n - 1} \right)\left( {3n - 4} \right)} \end{array}} \right|$,then the value of $\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} $:

If $A = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$,then the determinant of $A^2 - 2A$ is

Let $\alpha \in(0, \infty)$ and $A=\begin{bmatrix} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix}$. If $\operatorname{det}(\operatorname{adj}(2A-A^{T}) \cdot \operatorname{adj}(A-2A^{T}))=2^8$,then $(\operatorname{det}(A))^2$ is equal to: