If A= [ cc & - & ], then verify that A′ A = I

Question

If $A=\left[\begin{array}{cc} {\sin \alpha} & {\cos \alpha} \\ {-\cos \alpha} & {\sin \alpha} \end{array}\right]$, then verify that A′ A = I

✓

Answer

Here, $A=\left[\begin{array}{cc} {\sin \alpha} & {\cos \alpha} \\ {-\cos \alpha} & {\sin \alpha} \end{array}\right]$
$\therefore$ $A^{\prime}=\left[\begin{array}{cc} {\sin \alpha} & {-\cos \alpha} \\ {\cos \alpha} & {\sin \alpha} \end{array}\right]$
$\Rightarrow$ $A A^{\prime}=\left[\begin{array}{cc} {\sin \alpha} & {\cos \alpha} \\ {-\cos \alpha} & {\sin \alpha} \end{array}\right] \times\left[\begin{array}{cc} {\sin \alpha} & {-\cos \alpha} \\ {\cos \alpha} & {\sin \alpha} \end{array}\right]$
$\Rightarrow~ \mathrm{AA}^{\prime}=\left[\begin{array}{ll} {\sin \alpha \times \sin \alpha+\cos \alpha \times \cos \alpha} & {\sin \alpha \times(-\cos \alpha)+\cos \alpha \times \sin \alpha} \\ {-\cos \alpha \times \sin \alpha+\sin \alpha \times \cos \alpha} & {-\cos \alpha \times(-\cos \alpha)+\sin \alpha \times \sin \alpha} \end{array}\right]$
$\Rightarrow \mathrm{AA}^{\prime}=\left[\begin{array}{cc} {\sin ^{2} \alpha+\cos ^{2} \alpha} & {-\sin \alpha \cos \alpha+\cos \alpha \sin \alpha} \\ {-\cos \alpha \sin \alpha+\sin \alpha \cos \alpha} & {\cos ^{2} \alpha+\sin ^{2} \alpha} \end{array}\right]$
$\Rightarrow \mathrm{AA}^{\prime}=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right] = I \quad\left(\because \cos ^{2} \alpha+\sin ^{2} \alpha=1\right)$
Therefore, AA' = I
Hence verified.