Question

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

Answer 1

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`