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This question contains three parts. I have already answered the first two. The last part is confusing me.

Suppose $A$ is a $4 \times 4$ matrix whose characteristic polynomial is $p(x) = (x - 1)(x + 2)^2(x - 3)$.

Part (a): Show that $A$ is invertible. Find the characteristic polynomial of $A^{-1}$.

We have that the roots of a characteristic polynomial are the eigenvalues of $A$. That is, $\lambda = -2, -2, 1, 3$ are our eigenvalues. The determinant of an $n \times n$ matrix is the product of its eigenvalues. Hence, det$A = 12$. An $n \times n$ matrix is invertible if and only if its determinant is nonzero. Therefore, $A$ is invertible.

Since none of the eigenvalues are zero, we have that $\lambda$ is an eigenvalue of $A$ if and only if $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$. Then, the characteristic polynomial for $A^{-1}$ is $q(x) = (x - 1)(x + 1/2)^2(x - 1/3)$.

Part (b): Find the determinant and trace of $A$ and $A^{-1}$.

This is easy since the determinant of an $n \times n$ matrix is the product of its eigenvalues and the trace of an $n \times n$ matrix is the sum of its eigenvalues.

Part (c): Express $A^{-1}$ as a polynomial in $A$. Explain your answer.

Not really sure what part (c) is getting at.

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    $\begingroup$ For the part $a$, you did not address the issue of multiplicities... I suggest instead to note that $\det(xI-A)=\det(A)\det(xA^{-1}-I)=\det(A)*(-1/x)^4\det(x^{-1}I-A)$, which does give your result. $\endgroup$ – Mindlack Jan 6 at 0:49
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By the Cayley-Hamilton theorem, we have $(A-1)(A+2)^2(A-3)=0$, that is, $A^4-9A^2-4A+12I=0$. Multiply both sides by $A^{-1}$, and be amazed!

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    $\begingroup$ suggest you type in the $I$ for $12 I$ I see, you also wrote the factored version without any $I$ $\endgroup$ – Will Jagy Jan 6 at 0:49
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    $\begingroup$ You are right, of course. Thanks. $\endgroup$ – A. Pongrácz Jan 6 at 1:00
  • $\begingroup$ Much appreciated. Is there an approach that does not use Cayley-Hamilton? Or is this the "obvious" way to go about this problem? $\endgroup$ – Taylor McMillan Jan 6 at 1:07
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    $\begingroup$ Not sure about obvious, but certainly the simplest. $\endgroup$ – A. Pongrácz Jan 6 at 1:08
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Hint for part (c): Cayley-Hamilton. Then multiply by $A^{-1}$ and solve for the inverse.

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  • $\begingroup$ What tells you to use Cayley-Hamilton? Just the fact that it gives us a polynomial for $A$, and since $A$ is invertible, we should then be able to find a polynomial $A^{-1}$ in terms of $A$? $\endgroup$ – Taylor McMillan Jan 6 at 1:06

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