If $\left| \begin{matrix} x - 4 & 2x & 2x \\ 2x & x - 4 & 2x \\ 2x & 2x & x - 4 \end{matrix} \right| = (A + Bx)(x - A)^2$,then the ordered pair $(A, B) = $ . . . . .

If $|A|$ denotes the value of the determinant of the square matrix $A$ of order $3$,then $|-2A|=$

If ${a_1}, {a_2}, {a_3}, \dots, {a_n}, \dots$ are in $G$.$P$. and ${a_i} > 0$ for each $i$,then the value of the determinant $\Delta = \begin{vmatrix} \log {a_n} & \log {a_{n+2}} & \log {a_{n+4}} \\ \log {a_{n+6}} & \log {a_{n+8}} & \log {a_{n+10}} \\ \log {a_{n+12}} & \log {a_{n+14}} & \log {a_{n+16}} \end{vmatrix}$ is equal to

If $a, b,$ and $c$ are in $A$.$P$.,then the value of $\left|\begin{array}{lll}x+2 & x+3 & x+a \\ x+4 & x+5 & x+b \\ x+6 & x+7 & x+c\end{array}\right|$ is

The value of the determinant $\left| {\begin{array}{*{20}{c}}1&1&1\\{\cos (nx)}&{\cos (n + 1)x}&{\cos (n + 2)x}\\{\sin (nx)}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}} \right|$ is independent of:

$\left| {\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}} \right| = $