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Math Help - Orthogonal Matrix

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    Orthogonal Matrix

    Find the value of for which the matrix is orthogonal.
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  2. #2
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    Quote Originally Posted by mr_motivator View Post
    Find the value of for which the matrix is orthogonal.
    An Orthogonal matrix is one whos transpose is its inverse.

    In other words:

     AA^T = I = A^TA

    For your matrix, the transpose is:

     \left( \begin{array}{ccc}<br />
\frac{\sqrt{3}}{2} & \frac{1}{2} \\<br />
m & -\frac{\sqrt{3}}{2}<br />
  \end{array} \right)

    So multiply it by the original matrix
     \left( \begin{array}{ccc}<br />
\frac{\sqrt{3}}{2} & \frac{1}{2} \\<br />
m & -\frac{\sqrt{3}}{2}<br />
  \end{array} \right)\left( \begin{array}{ccc}<br />
\frac{\sqrt{3}}{2} & m \\<br />
\frac{1}{2}<br />
 & -\frac{\sqrt{3}}{2}<br />
  \end{array} \right)<br />
 = \left( \begin{array}{ccc}<br />
1 & 0 \\<br />
0<br />
 & 1<br />
  \end{array} \right)

     \left( \begin{array}{ccc}<br />
\bigg(\frac{\sqrt{3}}{2}\bigg)^2+\bigg(\frac{1}{2}  \bigg)^2 & \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} \\<br />
\frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4}  & m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2<br />
  \end{array} \right)<br />
 = \left( \begin{array}{ccc}<br />
1 & 0 \\<br />
0<br />
 & 1<br />
  \end{array} \right)

     \left( \begin{array}{ccc}<br />
1 & \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} \\<br />
\frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} & m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2<br />
  \end{array} \right)<br />
 = \left( \begin{array}{ccc}<br />
1 & 0 \\<br />
0<br />
 & 1<br />
  \end{array} \right)

    Hence  \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} =0 and  m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2 = 1
    Last edited by Mush; February 17th 2009 at 05:06 AM.
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