# Prove that $A^{-1}=-\frac{1}{c_0}(A^{n-1}+c_{n-1}A^{n-2}+…+c_1I_n)$

Assume A is an $n \times n$ invertible matrix with the following characteristic polynomial: $$f(X)=X^n+c_{n-1}X^{n-1}+...+c_1X+c_0\;.$$ We wish to show $$A^{-1}=-\frac{1}{c_0}(A^{n-1}+c_{n-1}A^{n-2}+...+c_1I_n)\;.$$ I have shown that if $A$ is invertible then $c_0 \neq 0$. But I'm not sure how to go about getting the proper derivation given the knowledge that I already know. Any help is appreciated. Thanks in advance.

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@tkm Do you know about the Cayley - Hamilton Theorem? – user38268 Nov 29 '12 at 12:19

HINT: What do you get if you multiply $-\dfrac{1}{c_0}\left(A^{n-1}+c_{n-1}A^{n-2}+...+c_1I_n\right)$ by $A$?

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Brian's answer makes it clear that's the inverse of the matrix, now how to get there: reinforcing what you wrote, $\,A\,$ is invertible iff $\,c_0=\pm\det A\,$ is not zero, so we get the following matricial equalities:

$$A^n+c_{n-1}A^{n-1}+...+c_aA+c_0\cdot I=0\stackrel{\text{Multiply through by}\,\,A^{-1}}\Longrightarrow A^{n-1}+...+c_1\cdot I+c_0A^{-1}=0\Longrightarrow$$

$$c_0A^{-1}=-\left(A^{-n-1}+...+c_2A+c_1\cdot I\right)\Longrightarrow A^{-1}=-\frac{1}{c_0}\left(A^{-n-1}+...+c_2A+c_1\cdot I\right)$$

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