Question

Professor Amartya Sen, an economist has presented a report on backward areas, stating that the backward regions have ample opportunities of their economic development if proper industries are established by the entrepreneurs seeing the availability of local resources. The unskilled and uneducated workforce can be employed in farming, mining, lumbering, hunting and fishing industries.
Identify the category and further sub division of the industry referred to in the above para.​

Answer 1

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