Can anyone explain why the 3rd operation applied on a system creates an equivalent system with the same solution.

Elementary Row Operations.

1. Interchange two rows.

2. Multiply a row with a nonzero number.

3. Add a row to another one multiplied by a number.

  • $\begingroup$ Replace the word "row" with "linear equation". $\endgroup$
    – Paul
    Jun 24, 2016 at 20:59

2 Answers 2


Think about it as if you are dealing with a system of equations. For example, $$x+2y=0\\3x+4y=4$$When you multiply equation $1$ by a certain number, let's say $-3$, the equation becomes $-3x-6y=0$, which does not change any value of the variables.
Then add this new equation to the second equation, now they become $-2y=4$, and you can now solve for $x$ and $y$.
So it is basically the same process as you solve for a system of equations with multiple variables, this is just changed into the matrix form.


a system of linear equations is equivalent to an matrix equation of type $ Ax = a$, then the three operations indicated are the transformation of the matrix equation $ Ax = b$ in the matrix equation $PAx = Pa$ where $P$ is an invertible matrix and so the two matrix equation have the same solutions.

trasformation in step (1) $R_i$ to $R_j$: that is $P = (a_ {kl})$ where $a_{ij}=a_{ji}=a_ {kk}=1$ for all $k$, $i\not=k\not=j$ and $a_ {kl}=0$ for all others $k, l$.

in step (2) $R_i$ to $\lambda R_i$ with $\lambda\not=0$: that is $P = (a_ {kl})$ where $a_{ii}=\lambda$, $a_{jj}=1, j\not=i$ and $a_ {kl}=0$ for all others $k, l$.

in step (3) $R_i $ to $R_i+\lambda R_j$: that is $P = (a_ {kl})$ where $a_{ij}=\lambda$, $a_{kk}=1$, and $a_ {kl}=0$ for all others $k, l$.


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