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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!

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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).

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  • $\begingroup$ 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 $r_i \rightarrow r_i+cr_j$. $\endgroup$
    – rmp
    Oct 30, 2020 at 8:16
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    $\begingroup$ @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 $-1$. $\endgroup$
    – twosigma
    Oct 30, 2020 at 8:28
  • $\begingroup$ Thanks @twosigma! It cleared my confusion regarding that particular operation. $\endgroup$
    – rmp
    Oct 30, 2020 at 8:53

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