# A question about Elementary Row Operation: Add a Multiple of a Row to Another Row

The task is that I have to prove the following statement, using Linear Algebra arguments:

Given a matrix A, then: To perform an ERO (Elementary Row Operation) type 3 :

(c * R_i) + R_k --> R_k (i.e. replace a row k by adding c-multiple of row i to row k) is the same as replacing a row k by subtracting a multiple of some row from another row

I just don't know how to formally prove this statement, like how the arguments should look like.

By some inspections, I'm pretty sure that doing

(c * R_i) + R_k --> R_k

is the same as doing:

R_i - (d * R_k) --> R_k

where d can be positive or negative, but it must have opposite sign with c. I use an example as follows:

A = (2 1 3, 4 3 1)

Then if I want to add row 2 to row 1, say, instead of doing (1 * 4) + 2 --> 6 and so on, I do 4 - [ (-1) * 2 ] --> 6 instead. Thus c = 1 and d = -1 in this case. That's why I conclude that the coefficient d should always be the opposite sign with coefficient c.

Would someone help me on how to construct a formal proof of the statement? I know how to go about the examples, but I understand examples are never proofs >_<

Thank you very much ^_^

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adding $c\cdot R_i$ to $R_k$ and replacing the old $R_k$ by the result is the same as subtracting $-c\cdot R_i$ from $R_k$ and replacing the old $R_k$ by the result. Are you sure you stated the question correctly? The two gray operations are not the same. – Barbara Osofsky Feb 11 '13 at 5:54
@BarbaraOsofsky: Actually, the 2 gray operations are my interpretation of the phrase "replacing a row k by subtracting a multiple of some row from another row" of the statement, since I type out the statement word-by-word. So... am I having a misunderstanding about the statement? – Cecile Feb 11 '13 at 6:07
The two grayed statements are clearly not the same. Let $$A=\left[\begin{array}[c]{ccc}2&1&3 \\ 4&3&1 \end{array}\right].$$ Adding $2*R_1+R_2$ and putting the result in $R_2$ gives $$A= \left[\begin{array}[c]{ccc}2&1&3 \\ 8&5&7 \end{array}\right].$$ You cannot get the same result by multiplying row $k$ by a number $\ne 1$ and then adding the other row to it. The definition of ERO requires $i\ne k$. – Barbara Osofsky Feb 11 '13 at 18:26

Note: If $c$ is some non-zero scalar, then
• adding $cR_i$ to $R_k$ and replacing the original $R_{k\text{ old}}$ by $(R_k + cR_i)$
• subtracting $−c⋅R_i$ from $R_k$ and replacing the old $R_k$ by the result $R_k - (-cR_i)$.
Since...$R_k + cR_i = R_k - (-cR_i)$