# If we have a matrices $A=\begin{bmatrix}a & b \\c & d \end{bmatri Clifton Sanchez 2021-11-23 Answered If we have a matrices \[A=\begin{bmatrix}a & b \\c & d \end{bmatrix} \text{ and } e_{12}(\lambda)=\begin{bmatrix}1 & \lambda \\0 & 1\end{bmatrix}$
then by doing product $Ae_{12}(\lambda)=\begin{bmatrix}a & a\lambda+b \\c & c\lambda+d \end{bmatrix} \text{ and } e_{12}(\lambda)A=\begin{bmatrix}a+c\lambda & b+d\lambda \\c & d\end{bmatrix}$
we can interpret that right multiplication by $$\displaystyle{e}_{{{12}}}$$ to A gives a column-operation: add $$\displaystyle\lambda$$-times first column to the second column. In similar way, left multiplication by $$\displaystyle{e}_{{{12}}}{\left(\lambda\right)}$$ to A gives row-operation on A.
Is there any conceptual (not computational, if any) way to see that elementary row and column operations on a matrix can be expressed as multiplication by elementary matrices on left or right, accordingly?

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Froldigh
Here conceptual and computational ideas go hand in hand. We can see this by looking at the multiplication of A with $$\displaystyle{e}_{{{12}}}{\left(\lambda\right)}$$ in some detail.
We have
$\begin{bmatrix}1 & \lambda \\0 & 1 \end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}+\begin{bmatrix}0 & \lambda \\0 & 0 \end{bmatrix}=I+\lambda\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix}$
It is the position $$\displaystyle_{\left\lbrace{12}\right\rbrace}$$ of the blue marked 1 which determines selected row resp. column of A.
We obtain
$Ae_{12}(\lambda)=A(I+\lambda\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix})=A+\lambda\begin{bmatrix}0 & a \\0 & c \end{bmatrix}$
$e_{12}(\lambda)A=(I+\lambda\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix})A=A+\lambda\begin{bmatrix}c & d \\0 & 0 \end{bmatrix}$
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Louis Smith
Let A,B be two matrices of order n.
We can describe B as B=$$\displaystyle{\left({b}_{{1}},\ldots,{b}_{{n}}\right)}$$, where $$\displaystyle{b}_{{i}}$$ is its column i.
Notice that AB=$$\displaystyle{\left({A}{b}_{{1}},\ldots,{A}{b}_{{n}}\right)}$$
Now apply some column-operation on AB.
For example, let's say that its column i is multiplied by $$\displaystyle\lambda$$
So we obtain $$\displaystyle{\left({A}{b}_{{1}},\ldots,\lambda{A}{b}_{{i}},\ldots,{A}{b}_{{n}}\right)}$$
Notice that $$\displaystyle{\left({A}{b}_{{1}},\ldots,\lambda{A}{b}_{{i}},\ldots,{A}{b}_{{n}}\right)}={A}{\left({b}_{{1}},\ldots,\lambda{b}_{{i}},\ldots,{b}_{{n}}\right)}$$
So in order to apply some column-operation on AB, we can first apply it on B and then multiply the resulting matrix with A.
Can you see what happens if B=Id?
The same reasoning can be used for row-opperations.

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