- #1
tnutty
- 326
- 1
Say there are these systems of equations :
x - 2y + z = 0
2y - 8z = 8
-4x + 5y + 9z = -9
In matrix form, it can be represented like this :
--
[1 -2 1 0] < -- row 1
[0 2 -8 8] < -- row 2
[-4 5 9 -9] < -- row 3
When we do elementary row operations, say on row3 = row3 + 4*row1. That changes
row 3 to [0 -3 13 -9], if we substitute this for old row 3 we get a simpler matrix, buts
slightly modified. My question is that, when we do a row operation, how is the
resultant matrix essentially the same, i.e has the same solution set ?
I mean when we do row3 += 4*row1, does that not alter row3 to become a different
row with different solution ? Maybe I missed something in algebra class. Basically, I do
not see why row3 = row3 + 4*row1, has the same solution as just row3 ? I mean
don't we have to do this, row3 + 4*row1 = row4 + 4row1 ? How come, we are only adding
it to one side ?
x - 2y + z = 0
2y - 8z = 8
-4x + 5y + 9z = -9
In matrix form, it can be represented like this :
--
[1 -2 1 0] < -- row 1
[0 2 -8 8] < -- row 2
[-4 5 9 -9] < -- row 3
When we do elementary row operations, say on row3 = row3 + 4*row1. That changes
row 3 to [0 -3 13 -9], if we substitute this for old row 3 we get a simpler matrix, buts
slightly modified. My question is that, when we do a row operation, how is the
resultant matrix essentially the same, i.e has the same solution set ?
I mean when we do row3 += 4*row1, does that not alter row3 to become a different
row with different solution ? Maybe I missed something in algebra class. Basically, I do
not see why row3 = row3 + 4*row1, has the same solution as just row3 ? I mean
don't we have to do this, row3 + 4*row1 = row4 + 4row1 ? How come, we are only adding
it to one side ?