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Homework Help: The Matrices

  1. Oct 12, 2008 #1

    (a) Let [tex]A = \left(\begin{array}{ccc}1\\2\\-2\end{array}\right)[/tex], B = (0 3 -1)
    Find AB and BA or else explain why it cannot be done.

    (b) Let [tex]R_{\theta} = \left(\begin{array}{ccc}cos\theta&-sin\theta\\sin\theta&cos\theta\end{array}\right)[/tex], Find [tex]R_{\theta}R_{-\theta}.[/tex]

    3. The attempt at a solution

    (a) I believe that AB & BA cannot be computed because one of them is a row while the other one is a column. A column represents a vector whilst a row represents a point.

    The only way to find AB is to find B transpose aka BT and hence compute ABT.

    Is this the right answer for part (a) ? Thanks.
  2. jcsd
  3. Oct 13, 2008 #2


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    No, both AB and BA can be computed. One is a number and the other is a 3x3 matrix. Why?
  4. Oct 13, 2008 #3
    Oh I realized that! Thanks a lot. I managed to compute both AB and BA.
    :smile: :smile:

    But I don't understand part (b), we are required to find [tex]R_{\theta}R_{-\theta}[/tex].
    I reckon we need to multiply the matrix [tex]R_{\theta}[/tex] by [tex]R_{-\theta}[/tex].

    I don't understand, what is [tex]R_{-\theta}[/tex]?
  5. Oct 13, 2008 #4


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    [tex]R_{-\theta}[/tex] is what you get when you replace all the [tex]\theta[/tex] in the matrix with [tex]-\theta[/tex].
  6. Oct 13, 2008 #5
    [tex]R_{\theta} = \left(\begin{array}{ccc}cos\theta&-sin\theta\\sin\theta&cos\theta\end{array}\right)[/tex]

    [tex]R_{-\theta} = \left(\begin{array}{ccc}cos(-\theta)&-sin(-\theta)\\sin(-\theta)&cos(-\theta)\end{array}\right)[/tex]

    Now I need multiply the two in order to find [tex]R_{\theta}R_{-\theta}[/tex]

    For example to find the entry of the 1st row/1st column we are required to do the following;

    cos(θ) . cos(-θ) + -sin(θ) . sin(-θ)

    Is there a simplification for this? If so, how should I simplify it because each time I get a wrong answer…
  7. Oct 13, 2008 #6


    Staff: Mentor

    Before you multiply, simplify the entries in R_-theta. For example, what is cos (-theta)? sin(-theta)?
  8. Oct 13, 2008 #7
    I seriously have no idea as how do you simplify the entries in R.

    But if I had to guess I'd say cos(θ) . cos(-θ) simplifies to cos2(-θ)

    But the problem is that one of the θ's is positive while the other one is negative.
  9. Oct 13, 2008 #8


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    cos(-theta)=cos(theta). Look it up. What about sin(-theta)?
  10. Oct 13, 2008 #9
    To simplify the entries of [tex]R_{-\theta}[/tex], think about a point on the unit circle.

    First, how do the x and y coordinates relate to the sine and cosine functions?

    Secondly, pick a [tex]\theta[/tex]. Look at the x and y values corresponding to that [tex]\theta[/tex]. How do they relate to the x and y values corresponding to [tex]-\theta[/tex]?
  11. Oct 14, 2008 #10

    P(x,y) = (cos([tex]\theta[/tex]), sin([tex]\theta[/tex]))

    Is it: sin(-θ) = sin(θ) ?

    Does this mean that the entries in R simplifies into:

    [tex]\left(\begin{array}{ccc}cos\theta&-sin\theta\\sin\theta&cos\theta\end{array}\right)[/tex] ?
  12. Oct 14, 2008 #11


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    No, it doesn't! You had before
    [tex]\left(\begin{array}{cc} cos(-\theta) & -sin(-\theta) \\ sin(-\theta) & cos(-theta)\end{array}\right)[/tex]
    and you were told that [itex]cos(-\theta)= cos(\theta)[/itex] and [itex]sin(-\theta)= -sin(-\theta)[/itex]
    Put those into your matrix.
  13. Oct 14, 2008 #12

    [tex]R_{-\theta} = \left(\begin{array}{ccc}cos(\theta)&sin(-\theta)\\-sin(-\theta)&cos(\theta)\end{array}\right)[/tex]

    And to find [tex]R_{\theta}R_{-\theta}[/tex] I multiply the two:

    [tex]\left(\begin{array}{ccc}cos\theta&-sin\theta\\sin\theta&cos\theta\end{array}\right)[/tex] . [tex]\left(\begin{array}{ccc}cos(\theta)&sin(-\theta)\\-sin(-\theta)&cos(\theta)\end{array}\right)[/tex]

    It's a 2x2 matrix;[tex]\left(\begin{array}{ccc}a_{1 1}&a_{12}\\a_{21}&a_{22}\end{array}\right)[/tex] so;

    a11 = cos([tex]\theta[/tex]) . cos([tex]\theta[/tex]) + (-sin(-[tex]\theta[/tex]) . (-sin(-[tex]\theta[/tex]) => cos2θ + sin2(-θ)

    a12 = cos(θ) . (sin(-θ)) + (-sin(θ)) . cos(θ)

    Am I on the right track?
  14. Oct 14, 2008 #13


    Staff: Mentor

    There was a typo in HallsofIvy's post. sin(-theta) is not equal to -sin(-theta); it should have been sin(-theta) = -sin(theta).

    You've received a lot of help with this problem. I hope that you will be able to finish this one without needing any more clues.
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