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3 and 4 .Determinants and Matrices
medium
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
A
$x$
B
${x^3}$
C
$14 + {x^2}$
D
${x^5}$
Solution
(d) $\left| {\,\begin{array}{*{20}{c}}{4 + {x^2}}&{ – 6}&{ – 2}\\{ – 6}&{9 + {x^2}}&3\\{ – 2}&3&{1 + {x^2}}\end{array}} \right| = {x^4}(14 + {x^2})$ $ = x.{x^3}(14 + {x^2})$
Hence, the determinant is divisible by $x$,${x^3}$ and $(14 + {x^2})$,
but not divisible by ${x^5}$.
Standard 12
Mathematics