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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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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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    ARe you then saying that 3I -A is equivalent to A-1? A^2-3A+A=0, and then you could say A^2=A-3A as well. So then is A-1 = A^2? If so, then is A^2 a square matrix, or is it a square matrix squared?2013-01-24
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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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    Yes I did, well I think I did. I posted an answer below. Can you verify if it is correct?2012-09-20
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Solve the equation for $I$, then factor the other side.