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$$(5A^T)^{-1} = \begin{bmatrix} -3 & -1 \\ 5 & 2 \\ \end{bmatrix}$$

$$(A^T)^{-1} = \begin{bmatrix} -3/5 & -1/5 \\ 1 & 2/5 \\ \end{bmatrix}$$

$$A^{-1} = \begin{bmatrix} -3/5 & 1 \\ -1/5 & 2/5 \\ \end{bmatrix}$$

$$A = \begin{bmatrix} -2/5 & 1 \\ -1/5 & 3/5 \\ \end{bmatrix}$$

Is this how it is done? I'm not sure if $det$ is ever used when going from an inverted matrix to a normal matrix. I thought $AA^{-1}$ always resulted in an identity matrix, but it doesn't seem so here. At which step did I go wrong?

$$AA^{-1} = \begin{bmatrix} 1/25 & 0 \\ 0 & 1/25 \\ \end{bmatrix}$$

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How did you go from your $A^{-1}$ to $A$ in steps 3 and 4? Also, did you forget to divide a term by $5$ in step 2? –  Amzoti Oct 4 '13 at 13:02
    
$A_{11}$ became $-A_{22}$ and $A_{22}$ became $-A_{11}$. –  riista Oct 4 '13 at 13:08
    
I'm not sure where this division comes from. I divided both sides between steps 1 and 2. $-3$ became $-3/5$, $-1$ became $-1/5$, etc. –  riista Oct 4 '13 at 13:10
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Sorry that comment was incorrect on the missing 5 term, but wolframalpha.com/input/… –  Amzoti Oct 4 '13 at 13:13
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Did the above resolve your issues? Regards –  Amzoti Oct 4 '13 at 14:11

2 Answers 2

up vote 2 down vote accepted

We have $$(\lambda A)^{-1}=\frac{1}{\lambda}A^{-1}$$ so your second equality is wrong.

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actually you have to multiply by $5$ not divise by $5$ $(5A^T)^{-1} = \frac{1}{5}{(A^T)}^{-1} $

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This seems counterintuitive. I thought "whatever you do on the left side you do on the right side", re: basic fundamentals of algebra. Why is it divided? –  riista Oct 4 '13 at 12:36
    
I just tried using multiplication instead and no dice. I get a matrix of {25, 0 \\ 0, 25} for $AA^{-1}$. –  riista Oct 4 '13 at 12:52

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