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Math Help - [SOLVED] Nonsingular matrix

  1. #1
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    [SOLVED] Nonsingular matrix

    Show that the matrix \begin{pmatrix}cos@ & sin@\\-sin@ & cos@\end{pmatrix} is nonsingular and compute its inverse.

    (@ - theta)
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  2. #2
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    Hello.
    A Matrix B is non-singular if det(B) \neq 0.
    So in this case the determinant of your matrix is 1, that means it is non-singular.
    I usually find inverse using Gauss method.
     \begin{pmatrix}\cos x & \sin x & \vdots & 1 & 0\\-\sin x & \cos x & \vdots & 0 & 1\end{pmatrix}
    First to make
    a_{21} = 0 we do the transformation Row2 +\frac{\sin x}{\cos x}*Row1

    after doing that transformation it will be

    \begin{pmatrix}\cos x & \sin x & \vdots & 1 & 0\\0 & \frac{1}{\cos x} & \vdots & \frac{\sin x}{\cos x} & 1\end{pmatrix}

    now a12 should be 0. So we do the transformation Row1-\sin x*\cos x*Row2
    you'll have

    \begin{pmatrix}\cos x & 0 & \vdots  & \cos^2 x & -\sin x \cos x \\0 & \frac{1}{cos x} & \vdots & \frac{\sin x}{\cos x} & 1\end{pmatrix}

    now you divide the first row by \cos x and the second row with \frac{1}{\cos x}

    and you'll have:

    \begin{pmatrix}1 & 0 & \vdots & \cos x & -\sin x \\ 0 & 1 & \vdots & \sin x & \cos x \end{pmatrix}

    So the inverse of that matrix is
     A^{-1}= \begin{pmatrix}\cos x & -\sin x \\ \sin x & \cos x \end{pmatrix}


    You can check your results by multyplying A x A^{-1} if the result is the Identity matrix I, then it's done well.
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  3. #3
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    Thank you!

    Thank you. You helped me out so much and I love to learn how you can check your work. I'm taking an online class so my only instruction is the textbook, which isn't that great and lacks examples. Thanks again!
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