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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants2 Marks
Question
Find values of x for which $\left| {\begin{array}{*{20}{c}} 3&x \\ x&1 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 3&2 \\ 4&1 \end{array}} \right|$.
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Answer

${\left( {3 - x} \right)^2} = 3 - 8$

$3 - {x^2} = 3 - 8$

$ - {x^2} = - 8$

$x = \pm \sqrt 8 $

$x = \pm 2\sqrt 2 $

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