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Linear Algebra, Cramers Rule - Solving for unknowns in a matrix

  1. Mar 11, 2013 #1
    Hello everyone, I have a linear algebra question regarding Cramer's rule.

    1. The problem statement, all variables and given/known data

    Using Cramer's rule, solve for x' and y' in terms of x and y.

    [tex]
    \begin{cases}
    x = x' cos \theta - y' sin \theta\\
    y = x' sin \theta + y'cos \theta
    \end{cases}
    [/tex]

    2. Relevant equations

    ##sin^2 \theta + cos^2 \theta = 1 ##

    3. The attempt at a solution

    I need a matrix to start off, so I form a matrix based on the right-hand side of x and y. I'm assuming that x' and y' are just alternative ways of writing ##x_1## and ##x_2##.

    [tex]
    Let A =
    \begin{bmatrix}
    cos \theta & -sin \theta\\
    sin \theta & cos\theta
    \end{bmatrix}
    [/tex]
    I form two more matrices, ##A_1## and ##A_2##.

    [tex]
    Let A_1 =
    \begin{bmatrix}
    x & -sin \theta\\
    y & cos\theta
    \end{bmatrix}
    [/tex]
    [tex]
    Let A_2 =
    \begin{bmatrix}
    cos \theta & x\\
    sin \theta & y
    \end{bmatrix}
    [/tex]
    I then find ##det(A)##. I get ##cos^2 \theta + sin^2 \theta ## which is ##1##.
    ##det(A_1) = x cos \theta + y sin \theta##
    ##det(A_2) = y cos \theta - x sin \theta##
    Lastly, I need to find the value of ##x'## and ##y'##, using Cramer's Rule.

    [tex]
    x' = \frac{det(A_1)}{det A} = \frac{x cos \theta + y sin \theta}{1}\\
    y' = \frac{det(A_2)}{det A} = \frac{y cos \theta - x sin \theta}{1}
    [/tex]

    Can anyone tell me if I'm on the right track for this problem?
     
    Last edited: Mar 11, 2013
  2. jcsd
  3. Mar 11, 2013 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    A good working rule to learn is: check it for yourself, by substituting your proposed solution in the original equations to see if it all works out.
     
  4. Mar 12, 2013 #3
    Ray, I completely forgot about being able to check questions like these. I want to sincerely thank you for reminding me of this rule.
     
  5. Mar 12, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    By the way, the original equations,
    [itex]x= x'cos(\theta)- y'sin(\theta)[/itex]
    [itex]y= x'sin(\theta)+ y'cos(\theta)[/itex]

    give the (x, y) coordinates of point (x', y') after a rotation through angle [itex]\theta[/itex], clockwise, around the origin. Going from (x, y) back to (x', y') is just a rotation in the opposite direction. So

    [itex]x'= xcos(-\theta)- y sin(-\theta)= xcos(\theta)+ ysin(\theta)[/itex],
    [itex]y'= xsin(-\theta)+ ycos(-\theta)= -x sin(\theta)+ y cos(\theta)[/itex].
     
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