Determinants Matrices Question 45
Question: The value of the determinant $ \begin{vmatrix} {{\cos }^{2}}54{}^\circ & {{\cos }^{2}}36{}^\circ & \cot 135{}^\circ \\ {{\sin }^{2}}53{}^\circ & \cot 135{}^\circ & {{\sin }^{2}}37{}^\circ \\ \cot 135{}^\circ & cos^{2}25{}^\circ & {{\cos }^{2}}65{}^\circ \\ \end{vmatrix} $ is equal to
Options:
A) $ -2 $
B) $ -1 $
C) $ 0 $
D) $ 1 $
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Answer:
Correct Answer: C
Solution:
- [c] $ \Delta = \begin{vmatrix} {{\cos }^{2}}54{}^\circ & {{\cos }^{2}}36{}^\circ & \cot 135{}^\circ \\ {{\sin }^{2}}53{}^\circ & \cot 135{}^\circ & {{\sin }^{2}}37{}^\circ \\ \cot 135{}^\circ & {{\cos }^{2}}25{}^\circ & {{\cos }^{2}}65{}^\circ \\ \end{vmatrix} $
$ = \begin{vmatrix} {{\cos }^{2}}54{}^\circ & {{\sin }^{2}}54{}^\circ & -1 \\ {{\cos }^{2}}37{}^\circ & -1 & {{\sin }^{2}}37{}^\circ \\ -1 & {{\cos }^{2}}25{}^\circ & {{\sin }^{2}}25{}^\circ \\ \end{vmatrix} $
$ C_1\to C_1+C_2+C_3 $
$ = \begin{vmatrix} {{\cos }^{2}}54{}^\circ +{{\sin }^{2}}54{}^\circ -1 & {{\sin }^{2}}54{}^\circ & -1 \\ {{\cos }^{2}}37{}^\circ -1+{{\sin }^{2}}37{}^\circ & -1 & {{\sin }^{2}}37{}^\circ \\ -1+{{\cos }^{2}}25{}^\circ +{{\sin }^{2}}25{}^\circ & {{\cos }^{2}}25{}^\circ & {{\sin }^{2}}25{}^\circ \\ \end{vmatrix} $
$ = \begin{vmatrix} & \begin{matrix} 0 & {{\sin }^{2}}54{}^\circ -1 \\ \end{matrix} \\ & \begin{matrix} 0 & -1{{\sin }^{2}}37{}^\circ \\ \end{matrix} \\ & \begin{matrix} 0 & {{\cos }^{2}}25{}^\circ {{\sin }^{2}}25{}^\circ \\ \end{matrix} \\ \end{vmatrix} =0 $
BETA
