GATE | GATE CS 1996 | Question 31

The matrices

commute under multiplication
(A) if a = b or Θ = nπ, n an integer
(B) always
(C) never
(D) if a cosΘ = b sin Θ

Answer: (A)

Explanation: For Commutativity, AB = BA
 AB=  \begin{bmatrix} \cos{\theta} & -\sin{\theta}\\ \sin{\theta} & \cos{\theta}\\ \end{bmatrix} \begin{bmatrix} a&0\\ 0&b\\ \end{bmatrix} = \begin{bmatrix} a\cos{\theta} & -b\sin{\theta}\\ a\sin{\theta} & b\cos{\theta}\\ \end{bmatrix} \\\\ BA=  \begin{bmatrix} a&0\\ 0&b\\ \end{bmatrix} \begin{bmatrix} \cos{\theta} & -\sin{\theta}\\ \sin{\theta} & \cos{\theta}\\ \end{bmatrix} = \begin{bmatrix} a\cos{\theta} & -a\sin{\theta}\\ b\sin{\theta} & b\cos{\theta}\\ \end{bmatrix}
Since AB = BA
 \begin{bmatrix} a\cos{\theta} & -b\sin{\theta}\\ a\sin{\theta} & b\cos{\theta}\\ \end{bmatrix} = \begin{bmatrix} a\cos{\theta} & -a\sin{\theta}\\ b\sin{\theta} & b\cos{\theta}\\ \end{bmatrix}
On Comparing corresponding elements we get-
 a\sin{\theta} = b\sin{\theta}
The above holds when either, a=b or \sin{\theta} = 0 which is true when \theta = n\pi.
Therefore option (A) is correct.

This explanation is provided by Chirag Manwani.

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