Question 1118807: Solve using gauss elimination method
x+2y+2z=1; 2x+y+z=2; 3x+2y+2z=3; and y+z=0 Found 2 solutions by Fombitz, greenestamps:Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website! Multiply eq. 2 by 2 and subtract from eq. 1,
.
.
.
Subtract eq. 3 from eq. 1,
.
.
.
Multiply eq. 4 by 2 and subtract from eq. 1,
.
.
.
So,
and then from eq. 4,
With 4 equations and only 3 variables, there is a possibility that the system is over-constrained; there might not be a solution. Fortunately, it turns out that is not the case.
The beginning matrix for Gaussian elimination is
With the standard procedure for Gaussian elimination, the first step is to get a 1 in row 1 column 1 and 0s in the rest of column 1. The 1 in row 1 column 1 is already there; we can use it to get 0's in rows 2 and 3 of column 1.
Replace row 2 with (row 2 - 2*row 1); replace row 3 with (row3 - 3*row1):
I won't show the details of the next few steps. Clearly rows 2 and 3 are multiples of row 4. So if we move row 4 to be row 2 and use row 2 to simplify the last two rows, we get
Last, replace row 1 with (row1 - 2*row2), giving
The two non-zero rows then give us the solution:
x = 1; y+z = 0
This of course represents an infinite family of solutions. It is common to represent that family using a parameter:
