If $\left| {\begin{array}{*{20}{c}}{x - 4}&{2x}&{2x}\\{2x}&{x - 4}&{2x}\\{2x}&{2x}&{x - 4}\end{array}} \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2},$ then the ordered pair $\left( {A,B} \right) = $. . . . .

$\left| {\begin{array}{*{20}{c}}0&a&{ - b}\\{ - a}&0&c\\b&{ - c}&0\end{array}} \right| = $

Let $\omega $ be a complex number such that  $2\omega + 1 = z$ where $z = \sqrt { - 3} $ . If $\left| {\begin{array}{*{20}{c}}1&1&1\\1&{ - {\omega ^2} - 1}&{{\omega ^2}}\\1&{{\omega ^2}}&{{\omega ^7}}\end{array}} \right| = 3k$ then $k$ is equal to :

Prove that the determinant $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$ is independent of $\theta$

$\left| {\,\begin{array}{*{20}{c}}{bc}&{bc' + b'c}&{b'c'}\\{ca}&{ca' + c'a}&{c'a'}\\{ab}&{ab' + a'b}&{a'b'}\end{array}\,} \right|$ is equal to

Let $a ,b ,c $ be such that $b + c \ne 0$  if

$\left| {\begin{array}{*{20}{c}}a&{a + 1}&{a - 1}\\{ - b}&{b + 1}&{b - 1}\\c&{c - 1}&{c + 1}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{a + 1}&{b + 1}&{c - 1}\\{a - 1}&{b - 1}&{c + 1}\\{{{\left( { - 1} \right)}^{n + 2}} \cdot a}&{{{\left( { - 1} \right)}^{n + 1}} \cdot b}&{{{\left( { - 1} \right)}^n} \cdot c}\end{array}} \right| = 0$ then $n$ equals to