# Why define three elementary row operations? When one of them can be performed by the others?

I'm learning about linear algebra and in the course we've defined three "elementary row operations"

$$(1) \text{ Switching any two rows}$$ $$(2) \text{ Non-Zero scaling of any row}$$ $$(3) \text{ Adding a multiple of one row to a different row}$$

However it seems all these operations can be performed by the two simpler operations:

$$(1) \text{ Non-Zero scaling of any row}$$ $$(2) \text{ Adding any two rows}$$

I mean why even have switching rows as an operation in the first place when it can be performed by the addition and scaling of rows? It just seems redundant, why add an extra operation that can already be performed by the other operations?

Edit: Because someone in the comments asked how the swapping of rows could be performed with just non-zero scaling and adding rows. If you wanted to swap, say row $p$ with row $q$ you would:

$$(1) \text{ Add row } q \text{ to row } p$$ $$(2) \text{ Multiply row } p \text{ by } -1$$ $$(3) \text{ Add row } p \text{ to row } q$$ $$(4) \text{ Multiply row } q \text{ by } -1$$ $$(5) \text{ Add row } q \text{ to row } p$$ $$(6) \text{ Multiply row } p \text{ by } -1$$

• @Itay4 You scale a row, add it to another row, then scale the row by the inverse of the first scalar. Apr 25, 2017 at 18:32
• The point of the third operation is exactly this: to make things shorter. You may, in fact, work with only two operations. Apr 25, 2017 at 18:33
• @Marra I thought the point was to have the row operations be as simple as possible? Also why have adding a scaled multiple of a row instead of just the sum of two rows, when you could scale it and then add them? Apr 25, 2017 at 18:36
• @Marra: there may be more to this, from an algorithmic viewpoint. Using switching, all information can be stored in the matrix as-you-go. If you want to perform a switch using only (1) and (2), you need to store additional information.
– mlc
Apr 25, 2017 at 18:36
• @nathan56 if you want the theory to be strict (in the sense that it reduces everything to the mininum needed) then you are correct. But then, when using Gauss-Jordan's method do solve linear systems, the third operation is a tool that makes everything simpler. Also, as pointed out by the previous comment, this might be due do algorithmic reasons. Apr 25, 2017 at 18:40