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3 and 4 .Determinants and Matrices
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જો ${\left| {\,\begin{array}{*{20}{c}}4&1\\2&1\end{array}\,} \right|^2} = \left| {\,\begin{array}{*{20}{c}}3&2\\1&x\end{array}\,} \right| - \left| {\,\begin{array}{*{20}{c}}x&3\\{ - 2}&1\end{array}\,} \right|$ તો $x =$
A
$-14$
B
$2$
C
$6$
D
$7$
Solution
(c) $\left| {\,\begin{array}{*{20}{c}}4&1\\2&1\end{array}\,} \right|\,\left| {\,\begin{array}{*{20}{c}}4&1\\2&1\end{array}\,} \right| = \left| {\,\begin{array}{*{20}{c}}3&2\\1&x\end{array}\,} \right| – \left| {\,\begin{array}{*{20}{c}}x&3\\{ – 2}&1\end{array}\,} \right|$
$ = \left| {\,\begin{array}{*{20}{c}}{17}&9\\9&5\end{array}\,} \right| = (3x – 2) – (x + 6)$
==> $85 – 81 = 2x – 8$
==> $4 + 8 = 2x$
==> $x = 6$.
Standard 12
Mathematics