Question 1118807
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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.<br>
The beginning matrix for Gaussian elimination is<br>
{{{matrix(4,4,1,2,2,1,2,1,1,2,3,2,2,3,0,1,1,0)}}}<br>
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.<br>
Replace row 2 with (row 2 - 2*row 1); replace row 3 with (row3 - 3*row1):
{{{matrix(4,4,1,2,2,1,0,-3,-3,0,0,-4,-4,0,0,1,1,0)}}}<br>
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<br>
{{{matrix(4,4,1,2,2,1,0,1,1,0,0,0,0,0,0,0,0,0)}}}<br>
Last, replace row 1 with (row1 - 2*row2), giving<br>
{{{matrix(4,4,1,0,0,1,0,1,1,0,0,0,0,0,0,0,0,0)}}}<br>
The two non-zero rows then give us the solution:<br>
x = 1; y+z = 0<br>
This of course represents an infinite family of solutions.  It is common to represent that family using a parameter:<br>
x = 1; y = t, x = -t