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Give a 2x2 matrix with Det(A)=+/-1 that is not orthogonal.

  1. Sep 22, 2012 #1
    I attached the problem.
    I only need help with part b), I provided part a) just to remind you guys what a orthogonal matrix was.

    The only 2x2 matrix I can think of with a determinant of + or - 1 is something like

    √1 0
    0 1

    The determinant of this would be √1 = + or - 1

    The second part asks me to show that this matrix is not orthogonal thus A^-1≠A^T

    A^T=
    √1 0
    0 1


    To calculate A^-1 I set up the 2x2 matrix with the identify matrix on the right and reduce so the left becomes the I matrix.
    √1 0 1 0
    0 1 0 1


    1 0 1/√1 0
    0 1 0 1


    But when I do this I get A^-1 =
    1/√1 0
    0 1


    Is this still considered = to A^T =
    √1 0
    0 1
    ?
     

    Attached Files:

  2. jcsd
  3. Sep 22, 2012 #2

    Dick

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    Science Advisor
    Homework Helper

    √1 is 1. Try thinking of matrices with nonzero off-diagonal elements.
     
  4. Sep 22, 2012 #3
    A 2x2 orthogonal matrix must satisfy:
    [tex]
    \left(\begin{array}{cc}
    a & b \\

    c & d
    \end{array} \right) \cdot \left(\begin{array}{cc}
    a & c \\

    b & d
    \end{array}\right) = \left(\begin{array}{cc}
    a^2 + b^2 & a \, c + b \, d \\

    a \, c + b \, d & c^2 + d^2
    \end{array}\right) = \left(\begin{array}{cc}
    1 & 0 \\

    0 & 1
    \end{array}\right)
    [/tex]
    You can simply make it non-orthogonal if you make the off-diagonal element to be non-zero, for example:
    [tex]
    a \, c + b \, d = 1
    [/tex]
    Now, the determinant is:
    [tex]
    a \, d - b \, c = \pm1
    [/tex]
    Solve these two equations for a, and b, for example, and you will get a two-parameter family of matrices that are non-orthogonal, but with determinant ±1.
     
  5. Sep 22, 2012 #4

    Dick

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    You can write down a lot of simple examples without thinking half that hard.
     
    Last edited: Sep 23, 2012
  6. Sep 23, 2012 #5
    I'm not sure if I misinterpreted this problem when it says Det(A)=±1, does that mean the + and - have to be both satisfied?

    or can I have a matrix that's simply
    2 1
    1 1
     
  7. Sep 23, 2012 #6
    yes, you and I can. Please unquote the solution.
     
  8. Sep 23, 2012 #7
    yes.
     
  9. Sep 23, 2012 #8
    Oh will, I guess I was thinking too hard. Usually when they give ±, I assume that both conditions have to be met.

    Can someone verify if I proved part a right? I'm second guessing myself.

    Det(A^-1)=Det(A^t)
    Since Det(A^t)=Det(A)
    Det(A)=Det(A^-1)
    Det(A^-1)=1/Det(A)
    Thus
    Det(A)=1/Det(A)
    (Det(A))^2=1
    Det(A)=±1
     
  10. Sep 23, 2012 #9
    yes, it is correct.
     
  11. Sep 23, 2012 #10

    Mentallic

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    Homework Helper

    The [itex]\pm[/itex] operator is called plus or minus, not plus and minus :tongue:
     
  12. Sep 23, 2012 #11
    and imposing the conditions
    [tex]
    x = 1 \wedge x = -1
    [/tex]
    leads to a contradiction
    [tex]
    1 = -1
    [/tex]
    so, that interpretation doesn't make sense.
     
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