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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 $A^{-1} = 3I - A$.

3 Answers 3

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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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    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 $3I - A$.2012-09-20
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    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?2012-09-20
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    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.2012-09-20
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    Meaning they commute? Thank you very much Michael!2012-09-20
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    Michael one last question please, "Prove that a scalar c and A is an invertible matrix then the scalar product cA is invertible". My attempt: since A is invertible we have that AA^-1=I so if we multiply a scalar c we have that cAA^-1 = Ic thus cI=Ic making it invertible. Was that good or was I way off?2012-09-20
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    They do commute (but not every pair of commuting matrices are inverses i.e. $(2I)I = I(2I) = 2I \neq I$). With the second question, you're not quite there yet. You need to use the fact that $A$ has an inverse matrix to construct an inverse matrix for $cA$. That is, can you find a matrix $B$ such that $(cA)B = I$? Unlike your first question, you know that $A$ is invertible. Also, the question should say that the scalar $c$ is non-zero because $0A = O$ which is not invertible.2012-09-20
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    Hm I tried your advice but I am having trouble incorporating c into the proof because I still end up getting (cA)B = Ic which isn't invertible.2012-09-20
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    You've started with $AB = I$ and multiplied both by $c$, but you already know what $B$ is in this equation, it is $A^{-1}$. Instead start with the equation $(cA)B = I$ so that if there is a matrix $B$ which satisfies the equation, it is the inverse of $cA$. Can you rearrange the equation $(cA)B = I$ to determine $B$?2012-09-20
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    Okay after thinking hard this is what I got: (cA)B = I then [(1/B)A^-1)(cA)B] = [(1/B)B(cAA^-1) = I ? Is that good?2012-09-20
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    $B$ is a matrix, so $1/B$ doesn't make sense. Start with $(cA)B = I$. You know that $A$ has an inverse, so if you have $AC = D$ you can premultiply both sides to obtain $C = A^{-1}D$ (similarly, if we have $CA = D$ we can postmultiply by $A^{-1}$ to get $C = DA^{-1}$). Can you rearrange $(cA)B = I$ so that $A$ is on the left or on the right?2012-09-20
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    I am very raw at proofs, this is only my second week doing them so pls forgive my ignorance. Okay so I got that : (cA)B = I so if we premultiply we get that B = [((1/c)A^-1)I] so it follows [(cA)((1/c)A^-1] = [c(1/c)(AA^-1)] = I ..2012-09-20
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    Looks good. You have $B = \frac{1}{c}A^{-1}I = \frac{1}{c}A^{-1}$ so $(cA)^{-1} = \frac{1}{c}A^{-1}$.2012-09-20
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    Wow. Thank you so very much !!!2012-09-20
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    diimension, did you notice that what you did in the second comment to your answer is exactly what I said to do in my answer?2012-09-20
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    @GerryMyerson Wow, no I did not. I did not see your comment. freaky! Thanks for your input though even though I didn't see it.2012-09-20
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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.