3 and 4 .Determinants and Matrices
medium

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

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