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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 $a \cos\theta = b \sin\theta$
in Linear Algebra by Veteran (52.1k points)
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–1
Ans. a

2 Answers

+16 votes
Best answer
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 $\theta = {n\pi}.$
by Boss (33.8k points)
edited by
0
Pls Explain How it Matrix Multication Which Means Before the Multipication Step ?
0

Yes (A) is correct option, because $sin(\theta)$ will be 0, ​​if $\theta$=n*π, n is an integer.

0
What if we take a and b common from both columns respectively in first multiplication

And we take a and b common from both rows respectively in second multiplication

Then answer should be b. Why are we not doing this?
0

We can not take common from a row or column in matrix like in determinant. In matrix we have to take common from each cell of the matrix that is why your approach is wrong. See this link for better understanding.

https://math.stackexchange.com/questions/2616865/can-i-extract-common-factor-from-a-column-in-matrix

0
Thanks
+1 vote
by (171 points)
edited by
0
this link is not related to given question
0
Sorry,

changed the link have a look
Answer:

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