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

Question

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

✓

Answer

We know A’ can be calculated by taking the transpose of the given matrix A.
Therefore, $A^{\prime}=\left[\begin{array}{ll} {\cos \alpha} & {-\sin \alpha} \\ {\sin \alpha} & {\cos \alpha} \end{array}\right]$
Now multiply A and A’. So,
$A A^{\prime}=\left[\begin{array}{cc} {\cos \alpha} & {\sin \alpha} \\ {-\sin \alpha} & {\cos \alpha} \end{array}\right] \times\left[\begin{array}{cc} {\cos \alpha} & {-\sin \alpha} \\ {\sin \alpha} & {\cos \alpha} \end{array}\right]$
$\Rightarrow \mathrm{AA}^{\prime}=\left[\begin{array}{cc} {(\cos \alpha \times \cos \alpha)+(\sin \alpha) \times(\sin \alpha)} & {\cos \alpha \times(-\sin \alpha)+(\sin \alpha) \times \cos \alpha} \\ {-\sin \alpha \times \cos \alpha+\cos \alpha \times(\sin \alpha)} & {-\sin \alpha \times(-\sin \alpha)+\cos \alpha \times \cos \alpha} \end{array}\right]$
$\Rightarrow \mathrm{AA}^{\prime}=\left[\begin{array}{cc} {\cos ^{2} \alpha+\sin ^{2} \alpha} & {-\sin \alpha \cos \alpha+\cos \alpha \sin \alpha} \\ {-\sin \alpha \cos \alpha+\cos \alpha \sin \alpha} & {\sin ^{2} \alpha+\cos ^{2} \alpha} \end{array}\right]$
$\Rightarrow \mathrm{AA}^{\prime}=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right] ...(1) \quad\left(\because \cos \alpha^{2}+\sin \alpha^{2}=1\right)$
And we know ‘I’ represents an identity matrix
Therefore, $I=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right] $...(2)
From equation 1 & 2 we can say that
AA’ = I
AA’ = I. Hence verified.