How do I use Gaussian elimination to solve a system of equations?

1 Answer
Aug 16, 2015

Answer:

The goals of Gaussian elimination are to get #1#s in the main diagonal and #0#s in every position below the #1#s,

Then you can use back substitution to solve for one variable at a time.

Explanation:

EXAMPLE:

Use Gaussian elimination to solve the following system of equations.

#x+2y+3z=-7#
#2x-3y-5z=9#
#-6z-8y+z=-22#

Solution:

Set up an augmented matrix of the form.

#((1,2,3,|,-7),(2,3,-5,|,9),(-6,-8,1,|,22))#

Goal 1. Get a 1 in the upper left hand corner.

Already done.

Goal 2a: Get a zero under the 1 in the first column.

Multiply Row 1 by #-2# to get

#((-2,-4,-6,|,14))#

Add the result to Row 2 and place the result in Row 2.

We signify the operations as #-2R_2+R_1→R_2#.

#((1,2,3,|,-7),(2,3,-5,|,9),(-6,-8,1,|,22)) stackrel(-2R_1+R_2→R_2)(→) ((1,2,3,|,-7),(0,-7,-11,|,23),(-6,-8,1,|,22))#

Goal 2b: Get another zero in the first column.

To do this, we need the operation #6R_1+R_3→R_3#.

#((1,2,3,|,-7),(0,-7,-11,|,23),(-6,-8,1,|,22)) stackrel(6R_2+R_3→R_3)(→) ((1,2,3,|,-7),(0,-7,-11,|,23),(0,4,19,|,-64))#

Goal 2c. Get the remaining zero.

Multiply Row 2 by #-1/7#.

#((1,2,3,|,-7),(0,-7,-11,|,23),(0,4,19,|,-64)) stackrel(-(1/7)R_2 → R_2)(→) ((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,4,19,|,-64))#

Now use the operation #-4R_2+R_3 →R_3#.

#((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,4,19,|,-64)) stackrel(-4R_2+R_3 →R_3)(→) ((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,0,89/7,|,-356/7))#

Multiply the third row by #7/89#.

#((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,0,89/7,|,-356/7)) stackrel(7/89R_3 →R_3)(→) ((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,0,1,|,-4))#

Goal 3. Use back substitution to get the values of #x#, #y#, and #z#.

Goal 3a. Calculate #z#.

#z =-4#

Goal 3b. Calculate #y#.

#y+11/7z=-23/7#
#y-44/7=-23/7#
#y=44/7-23/7=21/7#

#y=3#

Goal 3c. Calculate x.

#x+2y+3z=-7#
#x+6-12=-7#
#x-6=-7#

#x=1#

The solution is #x=1,y=3,z=-4#