If $A$ and $B$ are $3 \times 3$ matrices such that $AB = A$ and $BA = B$,then

If $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$,then $A^3 - 4A^2 - 6A$ is equal to:

Let $\alpha$ and $\beta$ be real numbers. Consider a $3 \times 3$ matrix $A$ such that $A^2 = 3A + \alpha I$. If $A^4 = 21A + \beta I$,then:

Matrix $A_r = \begin{bmatrix} r & r-1 \\ r-1 & r \end{bmatrix}$ for $r = 1, 2, 3, \dots$. If $\sum_{r=1}^{109} |A_r| = (\sqrt{10})^k$,then $k = $ . . . . . . . Where $|A_r| = \det(A_r)$.

Let $A = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$,$B = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$,and $P = \begin{bmatrix} 0 & 1 & 0 \\ x & 0 & 0 \\ 0 & 0 & y \end{bmatrix}$ be an orthogonal matrix such that $B = PAP^{-1}$ holds. Then:

The number of positive integral solutions of the equation $\left| {\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^2}y}&{{x^2}z}\\{x{y^2}}&{{y^3} + 1}&{{y^2}z}\\{x{z^2}}&{y{z^2}}&{{z^3} + 1}\end{array}} \right| = 11$ is