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The matrices $\begin{bmatrix} \cos\theta && -\sin\theta \\ \sin\theta && \cos\theta \end{bmatrix}$ and $\begin{bmatrix} a && 0\\ 0&& b \end{bmatrix}$ commute under multiplication

  1. if $a=b \text{ or } \theta = n\pi, n$ an integer
  2. always
  3. never
  4. if $a \cos\theta = b \sin\theta$
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4 Answers

Best answer
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24 votes
Answer: $A$

​$\begin{bmatrix} \cos (\theta)&- \sin (\theta) \\ \sin (\theta)&\cos (\theta) \end{bmatrix}*\begin{bmatrix} a& 0 \\ 0&b \end{bmatrix}=\begin{bmatrix} a\cos (\theta)&{-b}\sin (\theta) \\a \sin (\theta)&b \cos (\theta) \end{bmatrix}$

and

​$\begin{bmatrix} a& 0 \\ 0&b \end{bmatrix}*\begin{bmatrix} \cos (\theta)&- \sin (\theta) \\ \sin (\theta)&\cos (\theta) \end{bmatrix}=\begin{bmatrix} a\cos (\theta)&{-a}\sin (\theta) \\b \sin (\theta)&b \cos (\theta) \end{bmatrix}$

The multiplication will commute if

$a \sin (\theta) = b \sin (\theta)$  or a = b or $\theta = {n\pi}.$
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for multiplication under commutative then its must follow the rule AB=BA

for multiplication , matrices must be identity matrix then it will follow commutative property.

for matrix1=  det(cosθ*cosθ+sinθ*sinθ=1)  for this θ=nπ. to become identity matrix.

for matrix2=det(a*b=1) for this a=b=1 because then it will become identity matrix.

so option A is correct.
Answer:

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