3 and 4 .Determinants and Matrices
normal

$\left| {\begin{array}{*{20}{c}}
{4 + {x^2}}&{ - 6}&{ - 2}\\
{ - 6}&{9 + {x^2}}&3\\
{ - 2}&3&{1 + {x^2}}
\end{array}} \right|$ $;(x\neq0)$ is not divisible by

A

$x$

B

$x^3$

C

$14+x^2$

D

$x^5$

Solution

$\left|\begin{array}{ccc}{4+x^{2}} & {-6} & {-2} \\ {-6} & {9+x^{2}} & {3} \\ {-2} & {3} & {1+x^{2}}\end{array}\right|=x\left(x^{3}\right)\left(14+x^{2}\right)$

Standard 12
Mathematics

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