If we have a matrices $$A=\begin{bmatrix} a & b\\ c & d\end{bmatrix} \hskip5mm \mbox{ and } \hskip5mm e_{12}(\lambda)= \begin{bmatrix} 1 & \lambda \\ 0 & 1 \end{bmatrix} $$ then by doing product $$ Ae_{12}(\lambda) = \begin{bmatrix} a & a\lambda + b\\ c & c\lambda + d \end{bmatrix} \hskip5mm \mbox{ and } \hskip5mm e_{12}(\lambda)A = \begin{bmatrix} a +c\lambda & b+d\lambda \\ c & d \end{bmatrix} $$ we can interpret that right multiplication by $e_{12}$ to $A$ gives a column-operation: add $\lambda$-times first column to the second column.

In similar way, left multiplication by $e_{12}(\lambda)$ to $A$ gives row-operation on $A$.

Question: 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?

  • 1
    $\begingroup$ It seems to me that it should be like this: $$ e_{12}(\lambda) A = \begin{bmatrix} a +c\lambda & b+d\lambda \\ c & d \end{bmatrix} \hskip5mm \mbox{ and } \hskip5mm Ae_{12}(\lambda) = \begin{bmatrix} a & a\lambda + b\\ c & c\lambda + d \end{bmatrix} $$ $\endgroup$
    – kabenyuk
    Commented Oct 4, 2021 at 14:56
  • $\begingroup$ Thanks; I edited it; sorry for the mistake. $\endgroup$ Commented Oct 4, 2021 at 15:20
  • $\begingroup$ You can search for 3Blue1Brown they have provided intuition behind matrix operations. $\endgroup$
    – user960916
    Commented Oct 4, 2021 at 15:50

2 Answers 2


Let $A,B$ be two matrices of order $n$.

We can describe $B$ as $B=(b_1,\ldots,b_n)$, where $b_i$ is its column $i$.

Notice that $AB=(Ab_1,\ldots,Ab_n)$.

Now apply some column-operation on $AB$.

For example, let's say that its column $i$ is multiplied by $\lambda$.

So we obtain $(Ab_1,\ldots,\lambda Ab_i,\ldots,Ab_n)$.

Notice that $(Ab_1,\ldots,\lambda Ab_i,\ldots,Ab_n)=A(b_1,\ldots,\lambda b_i,\ldots,b_n)$.

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.


Here conceptual and computational ideas go hand in hand. We can see this by looking at the multiplication of $A$ with $e_{12}(\lambda)$ in some detail.

We have \begin{align*} \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&\color{blue}{1}\\ 0&0 \end{bmatrix} \end{align*}

It is the position $_{12}$ of the blue marked $1$ which determines selected row resp. column of $A$.

We obtain \begin{align*} Ae_{12}(\lambda)&=A\left(I+\lambda \begin{bmatrix}0&\color{blue}{1}\\ 0&0 \end{bmatrix}\right) =A+\lambda\begin{bmatrix} 0&\color{blue}{a}\\ 0&\color{blue}{c} \end{bmatrix}\\ e_{12}(\lambda)A&=\left(I+\lambda \begin{bmatrix}0&\color{blue}{1}\\ 0&0 \end{bmatrix}\right)A =A+\lambda\begin{bmatrix} \color{blue}{c}&\color{blue}{d}\\ 0&0 \end{bmatrix} \end{align*}


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