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यदि $A =\left[\begin{array}{cc}0 & -\tan ^{\alpha} \\ \tan \frac{\alpha}{2} & 0\end{array}\right]$ तथा $I$ कोटि $2$ का एक तत्समक आव्यूह है। तो सिद्ध कीजिए कि $I+A=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
Solution
$L.H.S.$
$I+A$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0\end{array}\right]$
$\left[\begin{array}{cc}1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1\end{array}\right]$ ………. $(1)$
$R.H.S$ :
$(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
$=\left(\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0\end{array}\right]\right)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
$=\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right]\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
$=\left[\begin{array}{cc}\cos \alpha+\sin \alpha \tan \frac{\alpha}{2} & -\sin \alpha+\cos \alpha \tan \frac{\alpha}{2} \\ -\cos \alpha \tan \frac{\alpha}{2}+\sin \alpha & \sin \alpha \tan \frac{\alpha}{2}+\cos \alpha\end{array}\right]$ ………. $(2)$
$1-2 \sin ^2 \frac{\alpha}{2}+2 \sin \frac{\alpha}{2}-\cos \frac{\alpha}{2} \tan \frac{\alpha}{2} -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\left(2 \cos ^2 \frac{\alpha}{2}-1\right) \tan \frac{\alpha}{2}$
$-\left(2 \cos ^2 \frac{\alpha}{2}-1\right) \tan \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \tan \frac{\alpha}{2}+1-2 \sin ^2 \frac{\alpha}{2}$
$1-2 \sin ^2 \frac{\alpha}{2}+2 \sin ^2 \frac{\alpha}{2} 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}-\tan \frac{\alpha}{2}$
$-2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\tan \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} 2 \sin ^2 \frac{\alpha}{2}+1-2 \sin ^2 \frac{\alpha}{2}$
$=\left[\begin{array}{cc}1 -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} 1\end{array}\right]$
Thus, from $( 1 )$ and $( 2 )$, we get $L.H.S. = R.H.S.$
