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This question already has an answer here:

$A=\begin{pmatrix} 2&0&0\\ 1&3&0\\ -3&5&3 \end{pmatrix}.$

I can find the characteristic polynomial $C_a(A)= -A^3 + 8A^2-21A+18$.

How to continue from here to use the Cayley-Hamilton theorem to find the inverse matrix, $A^{-1}$?

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marked as duplicate by Jonas Meyer, Adam Hughes, ncmathsadist, Michael Albanese, Brad Aug 14 '14 at 19:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Since we know $p(A) = 0$, you can multiply by $A^{-1}$ both sides to have:

$$ -A^2 + 8 A - 21 \, I+ 18 A^{-1} = 0,$$

can you solve for $A^{-1}$?

Spoiler:

The answer is given by $A^{-1} = \frac{1}{18} \left( 21 I - 8 A + A^2 \right) =\left(\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -\frac{1}{6} & \frac{1}{3} & 0 \\ \frac{7}{9} & -\frac{5}{9} & \frac{1}{3}\end{array}\right)$

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${Hint:}$

If $-A^3+8A^2-21A+18I=0$, then $A(-A^2+8A-21I)=-18I$.

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    $\begingroup$ An advantage of this approach, rather than multiplying by $A^{-1}$, is that one needs not know that $A^{-1}$ exists a priori. $\endgroup$ – Jonas Meyer Aug 14 '14 at 18:24
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    $\begingroup$ In this particular case, it's rather simple to show that $|A| \neq 0$, isn't it? $\endgroup$ – Dmoreno Aug 14 '14 at 19:40
  • $\begingroup$ @Dmoreno: It sure is. $\endgroup$ – Jonas Meyer Aug 14 '14 at 19:47

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