1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper

    [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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook