Question

If $A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$, $B=\left[\begin{array}{cc}\cos 2\beta & \sin 2\beta \\ \sin 2\beta & -\cos 2\beta \end{array}\right]$, where $0<\beta <\frac{\pi }{2}$, then prove that $BAB=A^{-1}$. .

Answer 1

We have $A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\Rightarrow A^{-1}=$$\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$.

Now, $BAB$

$=\left[\begin{array}{cc}\cos 2\beta & \sin 2\beta \\ \sin 2\beta & -cos2\beta \end{array}\right]$$\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$$\left[\begin{array}{cc}\cos 2\beta & \sin 2\beta \\ \sin 2\beta & -\cos 2\beta \end{array}\right]$

$=\left[\begin{array}{cc}\cos 2\beta & \sin 2\beta \\ \sin 2\beta & -\cos 2\beta \end{array}\right]$$\left[\begin{array}{cc}\cos (\alpha +2\beta )& \sin (\alpha +2\beta )\\ \sin (\alpha +2\beta )& -\cos (\alpha +2\beta )\end{array}\right]$

$=\left[\begin{array}{cc}\cos (\alpha +2\beta -2\beta )& \sin (\alpha +2\beta -2\beta )\\ -\sin (\alpha +2\beta -2\beta )& \cos (\alpha +2\beta -2\beta )\end{array}\right]$

$=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]=A^{-1}$.