Ajay has done a life insurance policy with LIC but he did not disclosed that he is suffering from TV hair which values have been affected by the act of Ajay question mark and what can be the consequences at the time of claiming policy a short money when he die soon?
Answers
Answer:
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▪Given :-
\begin{gathered} A = \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}\end{gathered}A=[cosθ−sinθsinθcosθ]
And
B=A+A^4B=A+A4
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▪To Calculate :-
det(B)
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▪Solution :-
\begin{gathered} A = \begin{bmatrix} \sf cos \theta& \sf sin \theta \\ \sf - sin \theta& \sf cos \theta \end{bmatrix}\end{gathered}A=[cosθ−sinθsinθcosθ]
So,
\begin{gathered} \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} \end{gathered}A2=[cosθ−sinθsinθcosθ]=[cosθ−sinθsinθcosθ][cosθ−sinθsinθcosθ][cosθ−sinθsinθcosθ]=[cos2θ−sin2θ−sinθcosθ−sinθcosθsinθcosθ+sinθcosθ−sin2θ+cos2θ]=[cos2θ−sin2θsin2θcos2θ]
Similarly,
\begin{gathered}A {}^{4} = \begin{bmatrix} \sf cos 4\theta& \sf sin 4\theta \\ \sf - sin 4\theta& \sf cos4 \theta \end{bmatrix}\end{gathered}A4=[cos4θ−sin4θsin4θcos4θ]
As,
Given Matrix
B = A + A {}^{4}B=A+A4
So,
\begin{gathered} \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} \end{gathered}B=[cosθ−sinθsinθcosθ]+[cos4θ−sin4θsin4θcos4θ]=[cosθ+cos4θ−(sinθ+sin4θ)sinθ+sin4θcosθ+cos4θ]
\begin{gathered} \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)\end{gathered}∴det(B)=(cosθ+cos4θ)2+(sinθ+sin4θ)2=cos2θ+cos24θ+2cosθcos4θ+sin2θ+sin24θ+2sinθsin4θ=2+2cos(3θ)
\begin{gathered} \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} \: )}}}\end{gathered}So,atθ=5πdet(B)=2+2cos53π=4cos2(103π)=4(410−25)2 det(B)=4