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Find an elementary matrix E such that EA = B

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

$$B = \begin{bmatrix} -4 & 2 \\ 5 & 3 \end{bmatrix}$$

$$\begin{bmatrix} 2 & -1 \\ 5 & 3 \end{bmatrix}*E=\begin{bmatrix} -4 & 2 \\ 5 & 3 \end{bmatrix}$$

So the Elementary Matrix is an Identity Matrix that has one elementary row operation performed on it. Multiplying an Elementary Matrix by A should result in a matrix that is equivalent to having that elementary row operation performed onto A.

I can see that only the first row of A is modified to obtain B and I can tell that the first row of A is scaled by a value of -2.

Therefore the Elementary Matrix should be the Identity Matrix with the first row scaled by -2.$$\begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix}$$

However, $$\begin{bmatrix} 2 & -1 \\ 5 & 3 \end{bmatrix}*\begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix}≠\begin{bmatrix} -4 & 2 \\ 5 & 3 \end{bmatrix}$$

What am I doing incorrectly?

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  • $\begingroup$ Do you require AE=B or EA=B? you've implied both in your question. $\endgroup$ – Stephen Feb 19 '18 at 3:30
  • $\begingroup$ I made an error in regards to the order. This was pretty dumb. $\endgroup$ – mathguy Feb 19 '18 at 3:33
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Since the operation is $B=EA$ we have \begin{equation} \begin{pmatrix} -2 & 0\\0 & 1 \end{pmatrix}\begin{pmatrix} 2 & -1\\5 & 3 \end{pmatrix} = \begin{pmatrix} -4 & 2\\5 & 3 \end{pmatrix}. \end{equation} Your elementary matrix is correct but you meant to multiply it to $A$ on the left not on the right.

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I would look at what the question is asking versus what your calculations.

The question is asking to find a matrix $E$ (the elementary row operation matrix) such that $EA = B$. But in your attempt at the problem you try to find $E$ by solving the equation $AE=B$, which will get you a different solution.

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You can simply solve the below equation for $E$.

$$EA=B$$

That is,

$$EA=B \Longrightarrow EAA^{-1}=BA^{-1} \Longrightarrow E=BA^{-1} .$$

So all you need is to find the inverse of $A$, and you're done.

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  • $\begingroup$ What if A does not have an inverse? Moreover, this method is not necessarely easy to implement. $\endgroup$ – Martigan Feb 19 '18 at 9:59
  • $\begingroup$ @Martigan I understand, but I was giving a method for this particular case. $A$ is invertible and it's straightforward to find the inverse of a 2x2 matrix, hence my suggestion. $\endgroup$ – Stephen Feb 21 '18 at 1:35

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