If $n$  be the number of values of $x$ for which
matrix $\Delta (x) =\left[ {\begin{array}{*{20}{c}}
{ - x}&x&2\\
2&x&{ - x}\\
x&{ - 2}&{ - x}
\end{array}} \right]$ will be singular, then $det(\Delta\,(n))$ is

$($ where $det(B)$ denotes determinant of Matrix $B) -$

  • A

    $-8$

  • B

    $-6$

  • C

    $0$

  • D

    $10$

Similar Questions

Evaluate $\Delta=\left|\begin{array}{ccc}0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0\end{array}\right|$

The determinant $\left| {\,\begin{array}{*{20}{c}}{4 + {x^2}}&{ - 6}&{ - 2}\\{ - 6}&{9 + {x^2}}&3\\{ - 2}&3&{1 + {x^2}}\end{array}\,} \right|$ is not divisible by

The value of $\left| {\,\begin{array}{*{20}{c}}{41}&{42}&{43}\\{44}&{45}&{46}\\{47}&{48}&{49}\end{array}\,} \right| = $

The value of $\left| {\begin{array}{*{20}{c}}
1&x&y\\
2&{\sin x + 2x}&{\sin y + 2y}\\
3&{\cos x + 3x}&{\cos y + 3y}
\end{array}} \right|$ is

The equation $\left| {\begin{array}{*{20}{c}}{{{(1 + x)}^2}}&{{{(1 - x)}^2}}&{ - \,(2 + {x^2})}\\{2x + 1}&{3x}&{1 - 5x}\\{x + 1}&{2x}&{2 - 3x}\end{array}} \right|$ $+$ $\left| {\begin{array}{*{20}{c}}{{{(1 + x)}^2}}&{2x + 1}&{x + 1}\\{{{(1 - x)}^2}}&{3x}&{2x}\\{1 - 2x}&{3x - 2}&{2x - 3}\end{array}} \right|$ $= 0$