If $C=2 \cos θ$, then the value of the determinant $Δ=\begin{vmatrix}C &1& 0\\1 &C& 1\\6&1& C\end{vmatrix}$, is
Answer & explanation
Correct answer: option 4
We have,
$Δ=\begin{vmatrix}C &1& 0\\1 &C& 1\\6&1& C\end{vmatrix}$
$⇒Δ=\begin{vmatrix}0 &1& 0\\1-C^2 &C& 1\\6-C&1& C\end{vmatrix}$ [Applying $C_1→C_1-CC_2$]
$⇒Δ=-\begin{vmatrix}1-C^2 &1\\6-C&C\end{vmatrix}$ [Expanding along $R_1$]
$⇒Δ=-(C-C^3-6+C)$
$⇒Δ=C^3 -2C +6$
$⇒Δ= 8 \cos^3 θ-4 \cos θ + 6$