Question

#Quality Question
@Matrices and Determinants

Let

$\theta = \frac{\pi}{5}$


And

$\sf A = \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}\ \textless \ br /\ \textgreater$


If B = A + A^4
then det(B) is

​

Answer 1

《¤¤¤¤¤¤¤¤¤¤¤¤¤¤》

▪Given :-

$A = \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}$

And

$B=A+A^4$

___________________________

▪To Calculate :-

det(B)

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▪Solution :-

$A = \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}$

So,

$\sf A {}^{2} = \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix} \\ \\ = \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix} \\ \\ \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix} \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix} \\ \\ = \small \begin{bmatrix} \sf cos {}^{2} \theta - {sin}^{2} \theta& \sf sin \theta cos \theta + sin \theta cos \theta \\ \sf - sin \theta cos \theta - sin \theta cos \theta& \sf - {sin}^{2} \theta + cos {}^{2} \theta \end{bmatrix} \\ \\ = \begin{bmatrix} \sf cos 2\theta& \sf sin 2\theta \\ \sf - sin2 \theta& \sf cos2 \theta \end{bmatrix}$

Similarly,

$A {}^{4} = \begin{bmatrix} \sf cos 4\theta& \sf sin 4\theta \\ \sf - sin 4\theta& \sf cos4 \theta \end{bmatrix}$

As,

Given Matrix

$B = A + A {}^{4}$

So,

$\sf B= \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}+ \begin{bmatrix} \sf cos 4\theta& \sf sin 4\theta \\ \sf - sin 4\theta& \sf cos4 \theta \end{bmatrix} \\ \\ = \begin{bmatrix} \sf cos \theta + cos 4\theta& \sf sin \theta + sin 4\theta \\ \sf -( sin \theta + sin 4\theta)& \sf cos \theta + cos4 \theta \end{bmatrix}$

$\bf \small\therefore det(B) = {(cos \theta + cos4 \theta)}^{2} + {(sin \theta + sin4 \theta)}^{2} \\ \\ = \sf {cos}^{2} \theta + {cos}^{2} 4\theta + 2 cos\theta cos4 \theta \\ + {sin}^{2} \theta \sf+ {sin}^{2} 4\theta + 2 sin\theta sin4 \theta \\ \\ = \sf 2 + 2cos(3 \theta)$

$\sf So, at \: \theta = \frac{\pi}{5} \\ \\ \sf det(B) = 2 + 2cos \frac{3\pi}{5} \\ \\ = \sf4 {cos}^{2} ( \frac{3\pi}{10} ) \\ \\ = \sf4(\frac{ \sqrt{10 - 2 \sqrt{5} } }{4} \: {)}^{2} \\ \\\large \colorbox{lime}{ \underline{\boxed{ \color{magenta}\bf det(B)= \frac{1}{4} (10 - 2 \sqrt{5} \: )}}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:A = \begin{gathered}\begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}\end{gathered}$

So,

$\rm :\longmapsto\: {A}^{2} = A \times A$

$\rm \:  =  \:  \:\begin{gathered}\begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}\end{gathered} \times \begin{gathered}\begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}\end{gathered}$

$\rm \:  =  \:  \:\bigg[ \begin{matrix} {cos}^{2}\theta - {sin}^{2}\theta &sin\theta cos\theta + cos\theta sin\theta \\ - sin\theta cos\theta - cos\theta sin\theta & {cos}^{2}\theta - {sin}^{2}\theta\end{matrix} \bigg]$

$\rm \:  =  \:  \:\bigg[ \begin{matrix} {cos}^{2}\theta - {sin}^{2}\theta &2sin\theta cos\theta \\ - 2sin\theta cos\theta & {cos}^{2}\theta - {sin}^{2}\theta\end{matrix} \bigg]$

We know that

$\boxed{ \sf{ \: {cos}^{2}x - {sin}^{2}x = cos2x}}$

and

$\boxed{ \sf{ \:2sinxcosx = sin2x}}$

$\rm \:  =  \:  \:\bigg[ \begin{matrix}cos2\theta &sin2\theta \\ - sin2\theta &cos2\theta \end{matrix} \bigg]$

Thus,

$\rm :\longmapsto\: {A}^{2} =  \:  \:\bigg[ \begin{matrix}cos2\theta &sin2\theta \\ - sin2\theta &cos2\theta \end{matrix} \bigg]$

Consider,

$\rm :\longmapsto\: {A}^{4} = {A}^{2} \times {A}^{2}$

$\rm \:  =  \:  \:\begin{gathered}\begin{bmatrix} \sf cos 2\theta& \sf sin 2\theta \\ \sf - sin2 \theta& \sf cos2 \theta \end{bmatrix}\end{gathered} \times \begin{gathered}\begin{bmatrix} \sf cos 2\theta& \sf sin 2\theta \\ \sf - sin 2\theta& \sf cos 2\theta \end{bmatrix}\end{gathered}$

$\rm \: =\bigg[ \begin{matrix} {cos}^{2}2\theta - {sin}^{2}2\theta &sin2\theta cos2\theta + cos2\theta sin2\theta \\ - sin2\theta cos2\theta - cos2\theta sin2\theta & {cos}^{2}2\theta - {sin}^{2}2\theta\end{matrix} \bigg]$

$\rm \:  =  \:  \:\bigg[ \begin{matrix}cos4\theta &sin4\theta \\ - sin4\theta &cos4\theta \end{matrix} \bigg]$

Thus,

$\rm :\longmapsto\: {A}^{4} =  \:  \:\bigg[ \begin{matrix}cos4\theta &sin4\theta \\ - sin4\theta &cos4\theta \end{matrix} \bigg]$

Now, It is further given that

$\rm :\longmapsto\:B = A + {A}^{4}$

$\rm :\longmapsto\:B = \begin{gathered}\begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}\end{gathered} + \:\bigg[ \begin{matrix}cos4\theta &sin4\theta \\ - sin4\theta &cos4\theta \end{matrix} \bigg]$

$\rm :\longmapsto\:B = \begin{gathered}\begin{bmatrix} \sf cos \theta + cos4\theta & \sf sin \theta + sin4\theta \\ \sf - sin \theta - sin4\theta & \sf cos \theta + cos4\theta \end{bmatrix}\end{gathered}$

So,

$\bf :\longmapsto\: |B|$

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

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

$\rm \:  =  \:  \:1 + 1 + 2(cos\theta cos4\theta + sin\theta sin4\theta )$

$\rm \:  =  \:  \:2 + 2cos(4\theta - \theta )$

$\rm \:  =  \:  \:2 + 2cos3\theta$

$\rm \:  =  \:  \:2(1 + cos3\theta)$

$\rm \:  =  \:  \:2 \times 2 {cos}^{2}\dfrac{3\theta }{2}$

$\rm \:  =  \:  \:2 \times 2 {cos}^{2}\dfrac{3\pi }{10} \: \: \: \: \: \: \: \: \{ \because \: \theta = \dfrac{\pi}{5} \}$

$\rm \:  =  \:  \: 4{cos}^{2}\dfrac{3\pi }{10} \:$

We know,

$\boxed{ \sf{ \:cos\dfrac{3\pi}{10} = \sqrt{\dfrac{10 - 2 \sqrt{5} }{16} }}}$

So,

$\boxed{ \sf{ \:cos^{2} \dfrac{3\pi}{10} = \dfrac{10 - 2 \sqrt{5} }{16} }}$

On using this value, we get

$\rm \:  =  \:  \:4 \times \dfrac{10 - 2 \sqrt{5} }{16}$

$\rm \:  =  \:  \: \dfrac{10 - 2 \sqrt{5} }{4}$

$\rm \:  =  \:  \: \dfrac{5 - \sqrt{5} }{2}$