Question

If A = $\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]$ and $A^{-1}$ = A', then find the value of $\alpha$.

Answer 1

We have, A = $\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]and{A}^{\prime }=\left[\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right]$

Also, $A^{-1}$ = A'

$\Rightarrow A{A}^{-1}$ = AA'

$\Rightarrow I=\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]\left[\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}co{s}^2\alpha +si{n}^2\alpha & 0\\ 0& si{n}^2\alpha +co{s}^2\alpha \end{array}\right]$

By using equality of matrices, we get

$co{s}^2\alpha +si{n}^2\alpha =1$

which is true for all real values of $\alpha$.