Question
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.
$$\frac{{2\,{{\sin }^2}2\theta }}{{\sin \theta }}$$
B.
$$8\,{\cos ^3}\theta - 4\cos \theta + 6$$
C.
$$\frac{{2\,{{\sin }}2\theta }}{{\sin \theta }}$$
D.
$$8\,{\cos ^3}\theta + 4\cos \theta + 6$$
Answer :
$$8\,{\cos ^3}\theta - 4\cos \theta + 6$$
Solution :
Given that,
\[\Delta = \left[ {\begin{array}{*{20}{c}}
C&1&0\\
1&C&1\\
6&1&C
\end{array}} \right] = C\left( {{C^2} - 1} \right) - 1\left( {C - 6} \right)\]
$$\eqalign{
& \Rightarrow \Delta = 2\cos \theta \left( {4\,{{\cos }^2}\theta - 1} \right) - \left( {2\cos \theta - 6} \right)\left( {\because C = 2\cos \theta {\text{ given}}} \right) \cr
& = 8\,{\cos ^3}\theta - 4\cos \theta + 6 \cr} $$