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Suppose $A = \begin{bmatrix}1&2&-1\\1&1&1\\1&-1&0\end{bmatrix}$ and $D = \begin{bmatrix}1&2&-1\\-3&-1&3\\2&1&-1\end{bmatrix}$. I need to find the matrix $E$ such that $EA = D$, if possible.

What I have tried is $E = D/A$. Would this yield the solution I am looking for as I'm not sure about how to solve this.

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    $\begingroup$ How about computing the inverse? $EA=D \rightarrow E=DA^{-1}$ $\endgroup$ – b00n heT Jan 22 '14 at 8:40
  • $\begingroup$ @b00nheT computing the inverse of $D/A$? Can you explain why I would do that? I was assuming in order to get E, I would just divide D by A. $\endgroup$ – Vanessa Jan 22 '14 at 8:41
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    $\begingroup$ I'm assuming you're not too familiar with matrices... Well.. To make it short there is no such thing as division in the "matrix realm" since a matrix is a linear application and hence to get to the identity it must be inverted... But you should start by getting the idea of what a matrix actually is before trying to solve exercises involving inverses... Try some basic literature of linear Algebra where you can find a well written introduction or try to grasp an idea by looking a the wikipedia article en.wikipedia.org/wiki/Matrix_(mathematics) $\endgroup$ – b00n heT Jan 22 '14 at 8:45
  • $\begingroup$ I'd say division doesn't even exist (not exactly) unless you look at it as multiplication by the inverse, which is how it should be looked at, at all times, in my opinion. $\endgroup$ – Git Gud Jan 22 '14 at 8:48
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Multiply $EA=D$ with $A^{-1}$ on both sides, you get

$$EA(A^{-1})=D(A^{-1})$$ So $E=D(A^{-1})$

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  • $\begingroup$ So $E$ = $\begin{bmatrix}1&0&0\\-2&1&-2\\1&0&1\end{bmatrix}$? $\endgroup$ – Vanessa Jan 22 '14 at 8:49
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    $\begingroup$ If you're not sure if you have the right result, you can always try and calculate $EA$ and see if you get $D$ or not. $\endgroup$ – 5xum Jan 22 '14 at 8:52
  • $\begingroup$ thanks a lot for making it clearer $\endgroup$ – Vanessa Jan 22 '14 at 8:59

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