Question

If $A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$, then $A{A}^T$ equals

• A. $\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]$
• B. $\left[\begin{array}{cc}{\cos }^2\theta & {\sin }^2\theta \\ {\sin }^2\theta & {\cos }^2\theta \end{array}\right]$
• C. $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Answer 1

Answer: (C)

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$A{A}^T=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]=\left[\begin{array}{cc}{\cos }^2\theta +{\sin }^2\theta & \cos \theta \sin \theta -\cos \theta \sin \theta \\ \cos \theta \sin \theta -\cos \theta \sin \theta & {\cos }^2\theta +{\sin }^2\theta \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
Ans: C