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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants2 Marks
Question
Prove that the determinant $\left| {\begin{array}{*{20}{c}} x&{\sin \theta }&{\cos \theta } \\ { - \sin \theta }&{ - x}&1 \\ {\cos \theta }&1&x \end{array}} \right|$ is independent of $\theta $.
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Answer

Let $\Delta = \left[ {\begin{array}{*{20}{c}} x&{\sin \theta }&{\cos \theta } \\ { - \sin \theta }&{ - x}&1 \\ {\cos \theta }&1&x \end{array}} \right]$Expanding along first row,
$\Delta = x\left| {\begin{array}{*{20}{c}} { - x}&1 \\ 1&x \end{array}} \right| - \sin \theta \left| {\begin{array}{*{20}{c}} { - \sin \theta }&1 \\ {\cos \theta }&x \end{array}} \right| $ $+ \cos \theta \left| {\begin{array}{*{20}{c}} { - \sin \theta }&{ - x} \\ {\cos \theta }&1 \end{array}} \right|$
$\Rightarrow \Delta = x\left( { - {x^2} - 1} \right) - \sin \theta \left( { - x\sin \theta - \cos \theta } \right) $ $+ \cos \theta \left( { - \sin \theta + x\cos \theta } \right)$
$\Rightarrow \Delta = - {x^3} - x + x{\sin ^2}\theta $ $+ \sin \theta \cos \theta - \sin \theta \cos \theta + x{\cos ^2}\theta$
$\Rightarrow \Delta = - {x^3} - x + x\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) $ $= -x^3 - x + x = - x^3$ which is independent of $\theta $

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