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) = $. . . . .
$\left( { - 4,3} \right)$
$\left( { - 4,5} \right)$
$\left( {4,5} \right)$
$\left( { - 4, - 5} \right)$
The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{10!}&{11!}&{12!}\\{11!}&{12!}&{13!}\\{12!}&{13!}&{14!}\end{array}\,} \right|$ is
If $A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right],$ find $|A|$.
If the system of linear equations $x - 2y + kz = 1$ ; $2x + y + z = 2$ ; $3x - y - kz = 3$ Has a solution $(x, y, z) \ne 0$, then $(x, y)$ lies on the straight line whose equation is
If the system of linear equations $2 x-3 y=\gamma+5$ ; $\alpha x+5 y=\beta+1$, where $\alpha, \beta, \gamma \in R$ has infinitely many solutions, then the value of $|9 \alpha+3 \beta+5 \gamma|$ is equal to
For $\alpha, \beta \in \mathrm{R}$ and a natural number $\mathrm{n}$, let
$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$