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) = $. . . . .

Let $\mathrm{A}$ be a square matrix of order $3 \times 3$ , then $|\mathrm{k A}|$ is equal to

The number of $\theta \in(0,4 \pi)$ for which the system of linear equations

$3(\sin 3 \theta) x-y+z=2$, $3(\cos 2 \theta) x+4 y+3 z=3$, $6 x+7 y+7 z=9$ has no solution is.

Let $M$ and $N$ be two $3 \times 3$ matrices such that $M N=N M$. Further, if $M \neq N^2$ and $M^2=N^4$, then

$(A)$ determinant of $\left( M ^2+ MN ^2\right)$ is $0$

$(B)$ there is a $3 \times 3$ non-zero matrix $U$ such that $\left( M ^2+ MN ^2\right) U$ is the zero matrix

$(C)$ determinant of $\left( M ^2+ MN ^2\right) \geq 1$

$(D)$ for a $3 \times 3$ matrix $U$, if $\left( M ^2+ MN ^2\right) U$ equals the zero matrix then $U$ is the zero matrix

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

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 $\theta$