If A = begin{bmatrix} 0 & -tan(frac{theta}{2} | vedclass

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(B) Let $t = 	an(\frac{	heta}{2})$. Then $A = \begin{bmatrix} 0 & -t \ t & 0 \end{bmatrix}$.$I_{2} + A = \begin{bmatrix} 1 & -t \ t & 1 \end{bmatr

Vedclass · Accepted Answer

$13$

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(B) Let $t = 	an(\frac{	heta}{2})$. Then $A = \begin{bmatrix} 0 & -t \ t & 0 \end{bmatrix}$.$I_{2} + A = \begin{bmatrix} 1 & -t \ t & 1 \end{bmatrix}$ and $I_{2} - A = \begin{bmatrix} 1 & t \ -t & 1 \end{bmatrix}$.$(I_{2} - A)^{-1} = \frac{1}{1 + t^{2}} \begin{bmatrix} 1 & -t \ t & 1 \end{bmatrix}$.$(I_{2} + A)(I_{2} - A)^{-1} = \frac{1}{1 + t^{2}} \begin{bmatrix} 1 & -t \ t & 1 \end{bmatrix} \begin{bmatrix} 1 & -t \ t & 1 \end{bmatrix} = \frac{1}{1 + t^{2}} \begin{bmatrix} 1 - t^{2} & -2t \ 2t & 1 - t^{2} \end{bmatrix} = \begin{bmatrix} \frac{1 - t^{2}}{1 + t^{2}} & -\frac{2t}{1 + t^{2}} \ \frac{2t}{1 + t^{2}} & \frac{1 - t^{2}}{1 + t^{2}} \end{bmatrix}$.Comparing with $\begin{bmatrix} a & -b \ b & a \end{bmatrix}$,we get $a = \frac{1 - t^{2}}{1 + t^{2}} = \cos 	heta$ and $b = \frac{2t}{1 + t^{2}} = \sin 	heta$.Thus,$a^{2} + b^{2} = \cos^{2} 	heta + \sin^{2} 	heta = 1$.Therefore,$13(a^{2} + b^{2}) = 13(1) = 13$.

If $A = \begin{bmatrix} 0 & -\tan(\frac{\theta}{2}) \\ \tan(\frac{\theta}{2}) & 0 \end{bmatrix}$ and $(I_{2} + A)(I_{2} - A)^{-1} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$,then $13(a^{2} + b^{2})$ is equal to ...........

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