Question

If A = $\left[\begin{array}{cc}\cos 2\mathrm{\theta }& \sin 2\mathrm{\theta }\\ -\sin 2\mathrm{\theta }& \cos 2\mathrm{\theta }\end{array}\right]$, find A².

Answer 1

$$\begin{gathered} \mathrm{Given}:A=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & cos2\theta \end{array}\right] \\ \mathrm{Now}, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}{\cos }^2\left(2\theta \right)-{\sin }^2\left(2\theta \right)& \cos \left(2\theta \right)\sin 2\theta +\cos \left(2\theta \right)\sin 2\theta \\ -\cos \left(2\theta \right)\sin 2\theta -\sin 2\theta \cos 2\theta & -{\sin }^2\left(2\theta \right)+{\cos }^2\left(2\theta \right)\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}\cos \left(2\times 2\theta \right)& 2\sin 2\theta \cos 2\theta \\ -2\sin 2\theta \cos \left(2\theta \right)& \cos \left(2\times 2\theta \right)\end{array}\right]\left[\because {\cos }^2\theta -{\sin }^2\theta =\cos 2\theta \right] \\ \Rightarrow A^2=\left[\begin{array}{cc}\cos 4\theta & \sin \left(2\times 2\theta \right)\\ -\sin \left(2\times 2\theta \right)& \cos 4\theta \end{array}\right]\left[\because \sin 2\theta =2\sin \theta \cos \theta \right] \\ \Rightarrow A^2=\left[\begin{array}{cc}\cos 4\theta & \sin 4\theta \\ -\sin 4\theta & \cos 4\theta \end{array}\right] \end{gathered}$$