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Lets say

$$ \text{let }A = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 3 & 1 \\ 4 & 1 & 2 \end{bmatrix} $$

Find a matrix C such that:

$$ \text{let }AC = \begin{bmatrix} 1 & 2 \\ 0 & 1 \\ 4 & 1 \end{bmatrix} $$

Let X be the matrix that AC equals too

$AC = X$

Is how you would solve this. $$A^{-1}AC = A^{-1}X$$

My question is, what side should $A^{-1}$ be? Left or right? I first assumed it doesn't matter but it clearly does in this question because if it was $XA^{-1}$ you would b e multiplying a 3x2 by 3x3 which isn't possible.

Another example:

C is a invertible matrix. D and A and C are also n x n matrices

$$D = C^{-1}AC$$

If I wanted D how would this look like? Is there two ways?

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The rule comes from $A=B\Rightarrow AC=BC$ and $DA=BA$ if their columns and rows meet. Hence your first argument is correct: to get $C,$ we have $A^{-1}AC=A^{-1}X$ and $C=A^{-1}X.$

For $D=C^{-1}AC,$ this is the simplest form you can get without other conditions. You cannot cancel $C^{-1}$ and $C$ for the multiplication of matrices is not commutative, i.e., generally $AB\neq BA.$(sometimes their columns and rows don't even meet!)

EDIT: If you want to get $A$ from $D=C^{-1}AC,$ likewise we have $$CDC^{-1}=CC^{-1}ACC^{-1}$$ and hence $$CDC^{-1}=A.$$ Just follow the side on which you multiply!

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  • $\begingroup$ $$CDC^{-1} = A$$ or $$C^{-1}DC = A$$. Sorry I meant A not D in my original question $\endgroup$ – user349557 Apr 10 '17 at 7:30
  • $\begingroup$ Ok, then I edited my answer. $\endgroup$ – tommy xu3 Apr 10 '17 at 7:32
  • $\begingroup$ Thx that helped me. $\endgroup$ – user349557 Apr 10 '17 at 7:44
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$A^{-1}$ has to be on the left, because if you right multiply both sides of the equation you get: $$ACA^{-1}=XA^{-1}$$ and you wouldn't simplify the $AA^{-1}$ part. Keep in mind that when you are dealing with matrix algebra you don't have the commutative property.

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