# Proof for Elementary Row Operations

The Elementary Row Operations are stated as follows:

1. Interchange two rows.
2. Multiply a row by a nonzero constant.
3. Add a multiple of a row to another row.

Consider the matrix below:

$$A =\begin{bmatrix}5 & 2 & 0 & 0 & -2\\0 & 1 & 4 & 3 & 2\\0 & 0 & 2 & 6 & 3\\0 & 0 & 3 & 4 & 1\\0 & 0 & 0 & 0 & 2 \end{bmatrix}$$

I am concerned with the third operation which is stated as $$r_i \rightarrow r_i+cr_j$$. Why is it wrong to interpret it as $$r_i \rightarrow cr_i+r_j$$? Is there any proof for this? Thank you!

There is nothing wrong with that interpretation. In that case you are simply taking the row $$r_i$$ and substituting it by the multiple $$cr_i$$ and then adding the row $$r_j$$.
Of course in this case you have to consider that $$c \neq 0$$, otherwise the matrix that you will obtain will not be equivalent to the previous one (i.e., all steps in row operations must be reversible).
• Thanks for this! But when I reduce a matrix using $r_i \rightarrow cr_i+r_j$ (into an upper triangular matrix), the determinant is not the same as when I do $r_i \rightarrow r_i+cr_j$. – rmp Oct 30 '20 at 8:16
• @rmp That's because they aren't the same operations. The operation $r_i \to cr_i + r_j$ multiplies the determinant of the original matrix by $c$. The operation $r_i \to r_i + cr_j$ doesn't change the determinant. In general, multiplying one row by a nonzero constant would multiply the determinant by that constant, whereas adding a multiple of one row to another row does not change the determinant, and as for the third operation of interchanging two rows, the determinant is multiplied by $-1$. – twosigma Oct 30 '20 at 8:28