Question

If $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$ then $A{A}^T$ is equal to: (where $A^T$ is the transpose of $A)$

• (A) Null matrix
• (B) Identify matrix
• (C) A
• (D) -A

Answer 1

The Correct Option is B

Explanation:
Given,$\begin{array}{r}A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\end{array}$The transpose of matrix $A$ is given by,$\begin{array}{r}{A}^T=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\end{array}$Now,$\begin{array}{rl}& A{A}^T=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\times \left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\\ & \Rightarrow A{A}^T=\left[\begin{array}{cc}{\cos }^2\alpha +{\sin }^2\alpha & -\cos \alpha \sin \alpha +\cos \alpha \sin \alpha \\ -\cos \alpha \sin \alpha +\cos \alpha \sin \alpha & {\sin }^2\alpha +{\cos }^2\alpha \end{array}\right]\\ & \Rightarrow A{A}^T=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\end{array}$$\Rightarrow A{A}^T=I$So, $A{A}^T$ is a identify matrix.Hence, the correct option is (B).