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Math Help - Matrix

  1. #1
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    Matrix

    A is a 3x3 matrix, and A^3 + A^2 + A + I = 0. How do I show that A has an inverse? Is it enough to show that A(-A^2 - A - 1) = I ?
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  2. #2
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    Here is one way that I might do it.
    A^3  + A^2  + A + I = 0\quad  \Rightarrow \quad A^4  + A^3  + A^2  + A = 0
    A^4  + A^3  + A^2  + A = 0\quad  \Rightarrow \quad A^4  + A^3  + A^2  + A + I = I
    A^4  + A^3  + A^2  + A + I = I\quad  \Rightarrow \quad A^4  + 0 = I
    A\left( {A^3 } \right) = \left( {A^3 } \right)A = I
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  3. #3
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    i don't think i'm quite follow what it is that you do..
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  4. #4
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    You are assuming:
    <br />
A^3 + A^2 + A + I = 0<br />
    So multiply both sides by the matrix A which gives you:
    <br />
A^4 + A^3 + A^2 + A = 0<br />
    Now add the identity matrix to both sides of the equation:
    <br />
A^4 + (A^3 + A^2 + A + I) = I<br />
    Observe that the bit in parantheses is the expression we assumed is exactly equal to the matrix of all 0's. Sustitute using that fact to arrive at:
    <br />
A^4 + 0 = I<br />
    The matrix A^4 + 0 is just A^4 which is really A^3(A) or (A)A^3.

    So not only have you shown that the inverse exists but also that the inverse of A is in fact A^3 since (A)A^3 = A^3(A) = I which is the definition of an inverse matrix.
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