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Using a piece from my last question I want to show how to find $A^{-1}$ as a polynomial expression in $A$ of degree < $\deg m_A$ where the leading coefficient of the polynomial is $\dfrac{-1}{m_A(0)}$ where $m_A$ is the minimal polynomial of $A$.

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if $f(x)$ is the minimal polynomial and $A$ is the matrix, what is $f(A)?$ – Will Jagy Dec 7 '12 at 1:45
$f(A) = 0$ correct? I am looking at a theorem in my book. – CodeKingPlusPlus Dec 7 '12 at 1:56
Yes. It is the lowest degree polynomial that does that. And it appears that you are calling the coefficient of $I$ in the minimal polynomial $m_A(0).$ Write it out, solve for $I,$ and see what happens. – Will Jagy Dec 7 '12 at 2:03
up vote 3 down vote accepted

If $m_A(x)=x^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0$ then since $p(A)=0\Rightarrow $ $$A^m+a_{m-1}A^{m-1}+\cdots+a_1A+a_0I=0\Rightarrow a_0I=-(A^m+a_{m-1}A^{m-1}+\cdots+a_1A)\Rightarrow\\ a_0I=A(-A^{m-1}-a_{m-1}A^{m-2}-\cdots-a_1)\Rightarrow \\I=A\left(-\dfrac{1}{a_0}A^{m-1}-\dfrac{a_{m-1}}{a_0}A^{m-2}-\cdots-\dfrac{a_1}{a_0}\right)\Rightarrow\\ A^{-1}=-\dfrac{1}{a_0}A^{m-1}-\dfrac{a_{m-1}}{a_0}A^{m-2}-\cdots-\dfrac{a_1}{a_0}.$$

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