Question

If A = $\left[\begin{array}{ccc}cos\alpha&sin\alpha\\-sin\alpha&cos\alpha\end{array}\right],$ show that AA' = A'A = I.

Answer 1

Answer:

AA' = A'A = I

Step-by-step explanation:

If

$A =\left[\begin{array}{ccc}cos\alpha&sin\alpha\\-sin\alpha&cos\alpha\end{array}\right]$

$A^{'} =\left[\begin{array}{ccc}cos\alpha&-sin\alpha\\sin\alpha&cos\alpha\end{array}\right]$

To show that AA' = A'A = I

$A.A^{'} =\left[\begin{array}{ccc}cos\alpha&sin\alpha\\-sin\alpha&cos\alpha\end{array}\right]\times\left[\begin{array}{ccc}cos\alpha&-sin\alpha\\sin\alpha&cos\alpha\end{array}\right]\\\\\\=\left[\begin{array}{ccc}cos^{2}\alpha+sin^{2}\alpha&sin\alpha\:cos\alpha-sin\alpha\:cos\alpha\\sin\alpha\:cos\alpha-sin\alpha\:cos\alpha&cos^{2}\alpha+sin^{2}\alpha\end{array}\right]\\\\\\=\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]\\\\A.A^{'}=I$

$A^{'}.A =\left[\begin{array}{ccc}cos\alpha&-sin\alpha\\sin\alpha&cos\alpha\end{array}\right]\times \left[\begin{array}{ccc}cos\alpha&sin\alpha\\-sin\alpha&cos\alpha\end{array}\right]\\\\\\=\left[\begin{array}{ccc}cos^{2}\alpha+sin^{2}\alpha&sin\alpha\:cos\alpha-sin\alpha\:cos\alpha\\sin\alpha\:cos\alpha-sin\alpha\:cos\alpha&cos^{2}\alpha+sin^{2}\alpha\end{array}\right]\\\\\\=\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]\\\\\\A^{'}.A=I$

so AA' = A'A = I