Let $A$ be a square matrix of order $3$ such that $\operatorname{det}(A)=-2$ and $\operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A)))=2^{m+n} \cdot 3^{mn}$,where $m > n$. Then $4m+2n$ is equal to . . . . . .

If $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \dots \begin{bmatrix} 1 & n-1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 78 \\ 0 & 1 \end{bmatrix}$,then the inverse of $\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$ is

Let $A$ be a $2 \times 2$ matrix.
$Statement-1: adj(adj A) = A$
$Statement-2: |adj A| = |A|$

If matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}$ and $A^{-1} = \alpha I + \beta A$ where $I$ is a unit matrix of order $2$ and $\alpha, \beta$ are constants,then the value of $\alpha + \beta + \alpha \beta$ is

If $A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{bmatrix}$,then $|\operatorname{adj} A|$ is equal to

If $\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} \right| = 5$; then the value of $\left| {\begin{array}{*{20}{c}}{{b_2}{c_3} - {b_3}{c_2}}&{{c_2}{a_3} - {c_3}{a_2}}&{{a_2}{b_3} - {a_3}{b_2}}\\{{b_3}{c_1} - {b_1}{c_3}}&{{c_3}{a_1} - {c_1}{a_3}}&{{a_3}{b_1} - {a_1}{b_3}}\\{{b_1}{c_2} - {b_2}{c_1}}&{{c_1}{a_2} - {c_2}{a_1}}&{{a_1}{b_2} - {a_2}{b_1}}\end{array}} \right|$ is: