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I am very raw at proofs, this is only my first semester learning them and I am having trouble with this problem. How would I approach this ?

Show that if a square matrix $A$ satisfies $A^2 - 3A + I = O$, then $A^{-1} = 3I - A$.

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Remember that a square matrix $B$ is the inverse of a square matrix $A$ if $AB = I$ (or $BA = I$; each one implies the other). Using the equation for $A$, can you show that $A(3I - A)$ or $(3I - A)A$ is equal to the identity matrix?

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  • $\begingroup$ Yes I did, well I think I did. I posted an answer below. Can you verify if it is correct? $\endgroup$
    – diimension
    Sep 20, 2012 at 5:07
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Solve the equation for $I$, then factor the other side.

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Actually nevermind, I got it! I will write the answer below for future reference.

If $A^2 - 3A + I = O,$ then, assuming $A^{-1}$ exists, multiply both sides of the equation by it: $$A^{-1}(A^2 - 3A + I) = A^{-1}O = O.$$ Expand the brackets: $$A - 3I + A^{-1} = O.$$ Leave $A^{-1}$ on the left: $$A^{-1} = 3I - A.$$

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  • $\begingroup$ As you state, you've assumed $A^{-1}$ exists. How do you know a square matrix $A$ satisfying $A^2 - 3A + I = O$ has an inverse? What you've shown is that if $A$ has an inverse and $A$ satisfies $A^2 - 3A + I = 0$, then its inverse is $3I - A$. $\endgroup$ Sep 20, 2012 at 5:08
  • $\begingroup$ A^2 - 3A + I = 0 then I = 3A - A^2 then I = A(3I - A) thus we conclude that A^(-1) = 3I - A. ? Will that satisfy the conditions you mentioned above? $\endgroup$
    – diimension
    Sep 20, 2012 at 5:11
  • $\begingroup$ Looks good to me (as long as you meant to put a comma between the $2$ and the $I$). Make sure that you can use the fact that (for square matrices) it is enough to show that $AB = I$, you may not have seen that result if this is homework; if you haven't seen the result, you can also rearrange the equation so that $A$ is on the right. $\endgroup$ Sep 20, 2012 at 5:19
  • $\begingroup$ Meaning they commute? Thank you very much Michael! $\endgroup$
    – diimension
    Sep 20, 2012 at 5:25
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    $\begingroup$ Wow. Thank you so very much !!! $\endgroup$
    – diimension
    Sep 20, 2012 at 6:48

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