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Question Number 9 by user1 last updated on 25/Jan/15

$$\mathrm{Let}A=\left[\begin{array}{cc}0& -\tan \frac{\mathrm{x}}{2}\\ \tan \frac{\mathrm{x}}{2}& 0\end{array}\right]\mathrm{and}I\mathrm{is}\mathrm{the}$$$$\mathrm{identity}\mathrm{matrix}\mathrm{of}\mathrm{order}2.\mathrm{Show}\mathrm{that}$$$$(I+A)=(I-A)\cdot \left[\begin{array}{cc}\cos x& -\sin x\\ \sin x& \cos x\end{array}\right].$$

Answered by user1 last updated on 30/Oct/14

$$\mathrm{Let}\tan \frac{x}{2}=t$$$$\mathrm{Then},\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\frac{1-t^2}{1+t^2}$$$$\mathrm{and}\sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\frac{2t}{1+t^2}$$$$\therefore (I+A)=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}0& -t\\ t& 0\end{array}\right]=\left[\begin{array}{cc}1& -t\\ t& 1\end{array}\right]$$$$(I-A)=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}0& -t\\ t& 0\end{array}\right]=\left[\begin{array}{cc}1& t\\ -t& 1\end{array}\right]$$$$\therefore (I-A)\cdot \left[\begin{array}{cc}\cos x& -\sin x\\ \sin x& \cos x\end{array}\right]$$$$=\left[\begin{array}{cc}1& t\\ -t& 1\end{array}\right]\left[\begin{array}{cc}\frac{1-t^2}{1+t^2}& \frac{-2t}{1+t^2}\\ \frac{2t}{1+t^2}& \frac{1-t^2}{1+t^2}\end{array}\right]$$$$=\left[\begin{array}{cc}\frac{1-t^2}{1+t^2}+\frac{2t^2}{1+t^2}& \frac{-2t}{1+t^2}+\frac{t(1-t^2)}{1+t^2}\\ \frac{-t(1-t^2)}{1+t^2}+\frac{2t}{1+t^2}& \frac{2t^2}{1+t^2}+\frac{1-t^2}{1+t^2}\end{array}\right]$$$$=\left[\begin{array}{cc}1& -t\\ t& 1\end{array}\right]=(I+A)$$$$\mathrm{Hence},(I+A)=(I-A)\left[\begin{array}{cc}\cos x& -\sin x\\ \sin x& \cos x\end{array}\right]$$

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