If $A = \begin{bmatrix} 0 & - an\frac{\alpha}{2} \ an\frac{\alpha}{2} & 0 \end{bmatrix}$ and $I$ is a 2$ imes 2$ unit matrix, then $(I - A)\begin{bmatrix} \cos\alpha & - \sin\alpha \ \sin\alpha & \sin\alpha \end{bmatrix}$ is

Question

If $A = \begin{bmatrix} 0 & - 	an\frac{\alpha}{2} \ 	an\frac{\alpha}{2} & 0 \end{bmatrix}$ and $I$ is a 2$	imes 2$ unit matrix, then $(I - A)\begin{bmatrix} \cos\alpha & - \sin\alpha \ \sin\alpha & \sin\alpha \end{bmatrix}$ is

Dr. Abubacker · Accepted Answer

(c)
Since $I = \begin{bmatrix}
1 & 0 \
0 & 1
\end{bmatrix} =$and given$\ \ A = \begin{bmatrix}
0 & 	an\frac{\alpha}{2} \
 - 	an\frac{\alpha}{2} & 0
\end{bmatrix}$
$	herefore I - A = \begin{bmatrix}
1 & - 	an\frac{\alpha}{2} \
	an\frac{\alpha}{2} & 1
\end{bmatrix}$(1)
Now, $(I - A)\begin{bmatrix}
\cos\alpha & - \sin\alpha \
\sin\alpha & \cos\alpha
\end{bmatrix}$
$= \begin{bmatrix}
1 & 	an\frac{\alpha}{2} \
 - 	an\frac{\alpha}{2} & 1
\end{bmatrix}\begin{bmatrix}
\cos\alpha & - \sin\alpha \
\sin\alpha & \cos\alpha
\end{bmatrix}$
$= \begin{bmatrix}
1 & 	an\frac{\alpha}{2} \
 - 	an\frac{\alpha}{2} & 1
\end{bmatrix}\begin{bmatrix}
\frac{1 - \frac{	an^{2}\alpha}{2}}{1 + \frac{	an^{2}\alpha}{2}} & - \frac{2	an\frac{\alpha}{2}}{1 + \frac{	an^{2}\alpha}{2}} \
\frac{2	an\frac{\alpha}{2}}{1 + \frac{	an^{2}\alpha}{2}} & \frac{1 - \frac{	an^{2}\alpha}{2}}{1 + \frac{	an^{2}\alpha}{2}}
\end{bmatrix}$
$= \left\lbrack \begin{matrix}
\frac{1 - \frac{	an^{2}\alpha}{2}}{1 + \frac{	an^{2}\alpha}{2}} + \frac{2	an^{2}\frac{\alpha}{2}}{1 + \frac{	an^{2}\alpha}{2}} \
\frac{- 	an\frac{\alpha}{2(1 - \frac{	an^{2}\alpha}{2})}}{1 + \frac{	an^{2}\alpha}{2}} + \frac{2	an\frac{\alpha}{2}}{1 + \frac{	an^{2}\alpha}{2}}
\end{matrix} ight.\$
$\left. \ \begin{matrix}
 - \frac{2	an^{}\frac{\alpha}{2}}{1 + \frac{	an^{2}\alpha}{2}} + \frac{	an\frac{\alpha}{2(1 - \frac{	an^{2}\alpha}{2})}}{1 + \frac{	an^{2}\alpha}{2}} \
\frac{2	an^{2}\frac{\alpha}{2}}{1 + \frac{	an^{2}\alpha}{2}} + \frac{1 - \frac{	an^{2}\alpha}{2}}{1 + \frac{	an^{2}\alpha}{2}}
\end{matrix} ightbrack$
$= \begin{bmatrix}
\frac{(1 + \frac{	an^{2}\alpha}{2})}{(1 + \frac{	an^{2}\alpha}{2})} & - \frac{	an\frac{\alpha}{2(1 + \frac{	an^{2}\alpha}{2})}}{(1 + \frac{	an^{2}\alpha}{2})} \
\frac{	an\frac{\alpha}{2(1 + \frac{	an^{2}\alpha}{2})}}{(1 + \frac{	an^{2}\alpha}{2})} & \frac{(1 + \frac{	an^{2}\alpha}{2})}{(1 + \frac{	an^{2}\alpha}{2})}
\end{bmatrix}$
$= \begin{bmatrix}
1 & - 	an\frac{\alpha}{2} \
	an\frac{\alpha}{2} & 1
\end{bmatrix}$
$= I - A\ \ \ \ \ \$[using (1)]

Q. If $A = \begin{bmatrix} 0 & - \tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 0 \end{bmatrix}$ and $I$ is a 2$\times 2$ unit matrix, then $(I - A)\begin{bmatrix} \cos\alpha & - \sin\alpha \\ \sin\alpha & \sin\alpha \end{bmatrix}$ is

Explanation

Reviewed by Dr. Abubacker

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