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Math Help - To find all invertible matrices of A

  1. #1
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    To find all invertible matrices of A

    I am asked to find all the invertible matrices of the form A=[(a, b) (c, d)] that satisfy A=A^-1 and A^t=a^-1 any cluues? I just know that the identity cos^2t+sin^2t=1 may be useful, cheeers!
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  2. #2
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    Re: To find all invertible matrices of A

    Hey Jordi1986.

    Hint: Consider the orthogonal matrices which have a determinant of 1 (look up rotation matrices if you are still confused).
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  3. #3
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    Re: To find all invertible matrices of A

    since A = A-1, and:

    det(A-1) = 1/det(A), we must have:

    det(A) = D = 1/det(A) = 1/D, so D2 = 1.

    thus D = det(A) = 1.

    now for:

    A = \begin{bmatrix}a&b\\c&d \end{bmatrix}

    we have:

    A^{-1} = \frac{1}{\det(A)}\begin{bmatrix}d&-b\\-c&a \end{bmatrix}}

    if det(A) = 1, this leads to b = c = 0, a = d = 1, that is: A = I (both of these matrices fit the bill).

    the case det(A) = -1 is more interesting:

    here, we can only conclude d = -a.

    however, we know by the orthogonality of A, that AT = A-1 = A, which means A must be symmetric, so we have b = c.

    thus our matrix is of the form:

    A = \begin{bmatrix}a&b\\b&-a \end{bmatrix}.

    since det(A) = -1 = -a2 - b2, it follows that:

    a2 + b2 = 1, or equivalently, b = √(1 - a2). thus A is of the form:

    \begin{bmatrix}a&\pm \sqrt{1-a^2}\\ \pm\sqrt{1-a^2}&-a \end{bmatrix} for a \in [-1,1] (the signs on the square root terms must match).

    if we so desired, we could write this as:

    A = \begin{bmatrix}\cos(\theta)&\sin(\theta)\\ \sin(\theta)&-\cos(\theta)\end{bmatrix} for 0 \leq \theta < 2\pi

    which is a REFLECTION about the line with slope: \tan\left(\frac{\theta}{2}\right).

    ignore what chiro said. there's only two rotations that work, but infinitely many reflections.
    Last edited by Deveno; February 12th 2013 at 12:56 AM.
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  4. #4
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    Re: To find all invertible matrices of A

    You're right I didn't see the A = A^(-1) [which does restrict the solution space a lot more].
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