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3 and 4 .Determinants and Matrices
medium
If $C = 2\cos \theta $, then the value of the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}C&1&0\\1&C&1\\6&1&C\end{array}\,} \right|$ is
A
$\frac{{\sin 4\theta }}{{\sin \theta }}$
B
$\frac{{2{{\sin }^2}2\theta }}{{\sin \theta }}$
C
$4{\cos ^2}\theta \,(2\cos \theta - 1)$
D
None of these
Solution
(d) $\Delta = \left| {\,\begin{array}{*{20}{c}}C&1&0\\1&C&1\\6&1&C\end{array}\,} \right|\, = C[{C^2} – 1] – 1[C – 6]$
==> $\Delta = 2\cos \theta (4{\cos ^2}\theta – 1) – (2\cos \theta – 6)$
==> $\Delta = 8{\cos ^3}\theta – 4\cos \theta + 6$.
Standard 12
Mathematics
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