GATE CSE 1996 | Question: 2.6

Question

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

• A. if $a=b \text{ or } \theta = n\pi, n$ an integer
• B. always
• C. never
• D. if $a \cos\theta = b \sin\theta$

Comments

Ans. a

Answer 1

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 $\theta = {n\pi}.$

Comments

Thanks

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What does commute mean exactly?

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It means AB = BA which is not true in general case. Also AB = I = BA.

Answer 2

https://www.youtube.com/watch?v=gWoqxrKmws4&t=2s

Comments

this link is not related to given question

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Sorry,

changed the link have a look

Answer 3

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 4

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