Question

Find the determinant of β?

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

Here

$A = \displaystyle\begin{pmatrix} cos \theta & sin \theta \\ - sin \theta & cos \theta \: \end{pmatrix}$

So

$\:{ A }^{2}$

$=A \times A$

$= \displaystyle\begin{pmatrix} cos \theta & sin \theta \\ - sin \theta & cos \theta \: \end{pmatrix} \times \displaystyle\begin{pmatrix} cos \theta & sin \theta \\ - sin \theta & cos \theta \: \end{pmatrix}$

$= \displaystyle\begin{pmatrix} {cos}^{2} \theta -{sin}^{2} \theta & 2sin \theta cos \theta\\ - 2sin \theta \: cos \theta & {cos}^{2} \theta -{sin}^{2} \theta \: \end{pmatrix}$

$= \displaystyle\begin{pmatrix} cos2 \theta & sin2 \theta \\ - sin2 \theta & cos 2\theta \: \end{pmatrix}$

Similarly

$\:{ A }^{3} = \displaystyle\begin{pmatrix} cos3 \theta & sin3 \theta \\ - sin3 \theta & cos 3\theta \: \end{pmatrix}$

$\:{ A }^{4} = \displaystyle\begin{pmatrix} cos4 \theta & sin4 \theta \\ - sin4 \theta & cos 4\theta \: \end{pmatrix}$

So

$\beta = \:{ A }^{4} + A$

$\implies \: \beta = \displaystyle\begin{pmatrix} cos4 \theta & sin4 \theta \\ - sin4 \theta & cos 4\theta \: \end{pmatrix} + \displaystyle\begin{pmatrix} cos \theta & sin \theta \\ - sin \theta & cos \theta \: \end{pmatrix}$

$\implies \: \beta = \displaystyle\begin{pmatrix} cos4 \theta +cos \theta & sin4 \theta + sin\theta \\ - (sin4\theta + sin\theta ) & cos4 \theta +cos \theta \: \end{pmatrix}$

Hence

$| \beta |$

$= \displaystyle\begin{vmatrix} cos4 \theta +cos \theta & sin4 \theta + sin\theta \\ - (sin4\theta + sin\theta ) & cos4 \theta +cos \theta \: \end{vmatrix}$

$= { \: (cos4 \theta +cos \theta )}^{2} + {( \:sin4\theta + sin\theta \: )}^{2}$

$= { \: {cos }^{2} 4 \theta +{cos}^{2} \theta } + { \: {sin }^{2} 4 \theta +{sin}^{2} \theta } + 2(cos4 \theta cos \theta + sin4 \theta sin\theta)$

$= { \: {cos }^{2} 4 \theta +{ \: {sin }^{2} 4 \theta +{cos}^{2} \theta } +{sin}^{2} \theta } + 2cos(4 \theta - \theta )$

$= 2 + 2cos3 \theta$