# Find Elementary Matrix

Find an elementary matrix E such that AE = B

$$A = \begin{bmatrix} 2 & 4 \\ 1 & 6 \end{bmatrix}$$

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

$$\begin{bmatrix} 2 & 4 \\ 1 & 6 \end{bmatrix}*E=\begin{bmatrix} 2 & -2 \\ 1 & 3 \end{bmatrix}$$

There should only be 1 elementary row operation performed on A. Scaling either row or swapping either row would not result in B, but I am unsure how to solve the specific pivot row operation that is used to obtain B.

• The only matrix $C$ such that $AC=B$ would be $C=A^{-1}B$, whether $C$ happened to be an elementary matrix or not. – JMoravitz Feb 19 '18 at 3:59
• Yes, I have the textbook answer (1 -3, 0 1) – mathguy Feb 19 '18 at 3:59
• @JMoravitz that method would not work for CA = B right? – mathguy Feb 19 '18 at 4:21
• If $CA=B$ then by right multiplication by $A^{-1}$ you get $C=BA^{-1}$. – JMoravitz Feb 19 '18 at 4:46

Since the matrix $E$ appears on the right of matrix $A$, think of elementary column operation.
Normally, one would use $E=A^{-1}B$ where $A^{-1}$ is the inverse of matrix $A$. Alternatively, we may let $$E=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ Hence one would have \begin{align} \begin{bmatrix} 2a+4c & 2b+4d \\ a+6c & b+6d \end{bmatrix}=\begin{bmatrix} 2 & -2 \\ 1 & 3 \end{bmatrix} \end{align} There rest should be easy peasy.
Think of getting $B$ from $A$ (it is a column operation): $$A=\begin{bmatrix}2 & 4 \\ 1 & 6 \end{bmatrix} \stackrel{-3\cdot C1+C2\to C2}{\Rightarrow} \begin{bmatrix}2 & -2 \\ 1 & 3 \end{bmatrix}=B.$$ Now do the same column operation with unit matrix: $$I=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \stackrel{-3\cdot C1+C2\to C2}{\Rightarrow} \begin{bmatrix}1 & -3 \\ 0 & 1 \end{bmatrix}=E.$$ Verify: $$AE=\begin{bmatrix}2 & 4 \\ 1 & 6 \end{bmatrix}\cdot \begin{bmatrix}1 & -3 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix}2 & -2 \\ 1 & 3 \end{bmatrix}=B.$$