Determinants Matrices Question 178
Question: If $ C=2cos\theta , $ then the value of the determinant $ \Delta = \begin{bmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \\ \end{bmatrix} $ is
Options:
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 $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Given that, $ \Delta = \begin{bmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \\ \end{bmatrix} =C(C^{2}-1)-1(C-6) $
$ \Rightarrow \Delta =2\cos \theta (4{{\cos }^{2}}\theta -1)-(2\cos \theta -6) $
$ (\because C=2\cos \theta given) $
$ =8{{\cos }^{3}}\theta -4\cos \theta +6 $
BETA
