Question

The matrices $\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$ and $\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]$ commute under multiplication

• A. if $a=b$ or $\theta =n\pi ,$ where $n$ is an integer
• B. always
• C. never
• D. if $a\cos \theta \ne b\sin \theta$

Answer 1

The correct option is A if $a=b$ or $\theta =n\pi ,$ where $n$ is an integer

$\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]=\left[\begin{array}{cc}a\cos \theta & -b\sin \theta \\ a\sin \theta & b\cos \theta \end{array}\right]$ ...(1)

And $\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]=\left[\begin{array}{cc}a\cos \theta & -a\sin \theta \\ b\sin \theta & b\cos \theta \end{array}\right]$ ...(2)

From (1) and (2), we get

$a\sin \theta =b\sin \theta \Rightarrow (a-b)\sin \theta =0$

either $a=b$ or $\sin \theta =0$

$\Rightarrow \theta =n\pi ;n\in Z$