Question

For $0<\theta <\pi$, if $A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$ , then

• A. $A^T=A$
• B. $A^T=-A$
• C. $A^2=I$
• D. $A^T=A^{-1}$

Answer 1

The correct option is D $A^T=A^{-1}$

$A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]A^T=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]A^{-1}=\frac{adjA}{\left|A\right|}\left|A\right|=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]={\cos }^2\theta +{\sin }^2\theta =1adjA=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]A^{-1}=\frac{adjA}{\left|A\right|}=\frac{\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]}{1}=A^{-1}\therefore A^T{=A}^{-1}$