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I'm working on the following problem and I can't seem to come up with the right answer. $$ \text{Let}: A^{-1} = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 3 & 1 \\ 4 & 1 & 2 \\ \end{bmatrix} $$ Find a matrix such that:

$$ ACA = \begin{bmatrix} 2 & 1 & 3 \\ -1 & 2 & 2 \\ 2 & 1 & 4 \\ \end{bmatrix} $$

Could someone point me in the right direction? Thanks!

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Your computation is wrong. You cannot just replace $AC$ with $(AC)^{-1}$. –  Martin Argerami Sep 15 '13 at 1:00
    
I don't see any instances of that in my question –  jollypianoman Sep 15 '13 at 1:01
    
oh nevermind now I do. –  jollypianoman Sep 15 '13 at 1:02
    
It is exactly what you did in your second equality: you wrote $ACA=(C^{-1}A^{-1})A$. Not true with the exception of some very particular choices of $A$ and $C$. –  Martin Argerami Sep 15 '13 at 1:02
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If you use bmatrix instead of matrix, you get the square brackets. –  Brian M. Scott Sep 15 '13 at 1:15

2 Answers 2

up vote 2 down vote accepted

Let $$ B = \begin{bmatrix} 2 & 1 & 3 \\ -1 & 2 & 2 \\ 2 & 1 & 4 \\ \end{bmatrix} $$

$ACA = B$ if and only if $A^{-1}ACA = A^{-1}B$ if and only if $A^{-1}ACAA^{-1} = A^{-1}BA^{-1}$. Now, multiplication between matrices is not commutative but it is associative! Hence you have: $$A^{-1}ACAA^{-1} = (A^{-1}A)C(AA^{-1}) = C$$

Then to find such a matrix $C$ you just need to calculate $A^{-1}BA^{-1}$, which is something you can do explicitly.

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You have $$ C=A^{-1}(ACA)A^{-1}. $$ So $$ C= \begin{bmatrix} 1 & 2 & 1 \\ 0 & 3 & 1 \\ 4 & 1 & 2 \\ \end{bmatrix}\,\cdot\, \begin{bmatrix} 2 & 1 & 3 \\ -1 & 2 & 2 \\ 2 & 1 & 4 \\ \end{bmatrix}\,\cdot\, \begin{bmatrix} 1 & 2 & 1 \\ 0 & 3 & 1 \\ 4 & 1 & 2 \\ \end{bmatrix}. $$ Now you can just perform the computation.

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