Using Gaussian Elimination on the given system, we start from
#{(2x-y+z=6),(x+2y-z=1),(2x-y-z=0):}#
and use the system to set up the augmented matrix
#((2,-1,1,|,6),(1,2,-1,|,1),(2,-1,-1,|,0))#
Our goal is a matrix of the form
#((1,0,0,|,x),(0,1,0,|,y),(0,0,1,|,z))#
Note that from here, we may use only the following operations:
-
multiply a row by a constant
-
add a the product of a row and a constant to another row
-
switch the positions of two rows
To obtain a #1# in the first row, first column, we multiply the first row by #1/2# and obtain
#((1,-1/2,1/2,|,3),(1,2,-1,|,1),(2,-1,-1,|,0))#
To obtain #0#s for the remainder of the first column, we add #-1# times the first row to the second row, and #-2# times the first row to the third row.
#((1,-1/2,1/2,|,3),(0,5/2,-3/2,|,-2),(0,0,-2,|,-6))#
We could solve for #z# and use back-substitution at this point, but let's proceed with Gauss-Jordan elimination. The next step is to get a #1# in the second row, second column. To do so, we multiply the second row by #2/5#.
#((1,-1/2,1/2,|,3),(0,1,-3/5,|,-4/5),(0,0,-2,|,-6))#
Again, we want #0#s for every other member of the column. To get this, we add #1/2# times the second row to the first row.
#((1,0,1/5,|,13/5),(0,1,-3/5,|,-4/5),(0,0,-2,|,-6))#
As our next to last step, we want a #1# in the third column, third row. To obtain this, we multiply the third row by #-1/2#.
#((1,0,1/5,|,13/5),(0,1,-3/5,|,-4/5),(0,0,1,|,3))#
Finally, we eliminate the remaining nonzero elements of the third column by adding #-1/5# times the third row to the first row, and #3/5# times the third row to the second row.
#((1,0,0,|,2),(0,1,0,|,1),(0,0,1,|,3))#
And so we have
#{(x = 2),(y=1),(z=3):}#
