Systems of EquationsGaussian Elimination

&

Row Reduced Echelon Form

by

Jeffrey Bivin

Lake Zurich High School

jeff.bivin@lz95.orgLast Updated: October 17, 2005

Example 1

Jeff Bivin -- LZHS

x + y + z = 64x – 8y + 4z = 122x – 3y + 4z = 3

1 1 1 6

4 -8 4 12

2 -3 4 3

Jeff Bivin -- LZHS

1 1 1 6

4 -8 4 12

2 -3 4 3

I am a 1.

Jeff Bivin -- LZHS

1 1 1 6

4 -8 4 12

2 -3 4 3

I need to be 0.

I need to be 0.

Jeff Bivin -- LZHS

1 1 1 6

0 -12 0 -12

0 -5 2 -9

4 - 4(1)-8 - 4(1)4 - 4(1)

12 - 4(6)

2 - 2(1)-3 - 2(1)4 - 2(1)3 - 2(6)

12 4RR

13 2RR

1 1 1 6

0 -12 0 -12

0 -5 2 -9

I need to be 1

Jeff Bivin -- LZHS

2121 R

1 1 1 6

0 1 0 1

0 -5 2 -9

1 1 1 6

0 1 0 1

0 -5 2 -9

I need to be 0.

I need to be 0.

Jeff Bivin -- LZHS

1 0 1 5

0 1 0 1

0 0 2 -4

1 - 01 - 11 - 06 - 1

0 + 5(0)-5 + 5(1)2 + 5(0)-9 + 5(1)

21 RR

23 5RR

1 0 1 5

0 1 0 1

0 0 2 -4

I need to be 1

Jeff Bivin -- LZHS

321 R

1 0 1 5

0 1 0 1

0 0 1 -2

1 0 1 5

0 1 0 1

0 0 1 -2

I need to be 0.

I am a 0

Jeff Bivin -- LZHS

1 0 0 7

0 1 0 1

0 0 1 -2

1 - 00 - 01 - 1

5 – (-2)

31 RR

1 0 0 7

0 1 0 1

0 0 1 -2

x = 7

y = 1

z = -2

Reading the Solution

Jeff Bivin -- LZHS

Writing the Solution

x + y + z = 6

4x – 8y + 4z = 12

2x – 3y + 4z = 3

Jeff Bivin -- LZHS

Example 2

Jeff Bivin -- LZHS

x + y + z = -2 2x - 3y + z = -11

-x + 2y - z = 8

1 1 1 -2

2 -3 1 -11

-1 2 -1 8

Jeff Bivin -- LZHS

1 1 1 -2

2 -3 1 -11

-1 2 -1 8

I am a 1.

Jeff Bivin -- LZHS

1 1 1 -2

2 -3 1 -11

-1 2 -1 8

I need to be 0.

I need to be 0.

Jeff Bivin -- LZHS

1 1 1 -2

0 -5 -1 -7

0 3 0 6

2 - 2(1)-3 - 2(1)1 - 2(1)

-11 - 2(-2)

-1 + 1 2 + 1-1 + 1

8 + (-2)

12 2RR

13 RR

1 1 1 -2

0 -5 -1 -7

0 3 0 6

I would prefer to make the 3 a one in row three rather than the -5

in row 2. Why?

Jeff Bivin -- LZHS

1 1 1 -2

0 3 0 6

0 -5 -1 -7

To avoid

fractions!

We will switch

Row 2 and Row 3

1 1 1 -2

0 3 0 6

0 -5 -1 -7

I need to be 1

Jeff Bivin -- LZHS

231 R

1 1 1 -2

0 1 0 2

0 -5 -1 -7

1 1 1 -2

0 1 0 2

0 -5 -1 -7

I need to be 0.

I need to be 0.

Jeff Bivin -- LZHS

1 0 1 -4

0 1 0 2

0 0 -1 3

1 - 01 - 11 - 0-2 - 2

0 + 5(0)-5 + 5(1)-1 + 5(0)-7 + 5(2)

21 RR

23 5RR

1 0 1 -4

0 1 0 2

0 0 -1 3

I need to be 1

Jeff Bivin -- LZHS

31R

1 0 1 -4

0 1 0 2

0 0 1 -3

1 0 1 -4

0 1 0 2

0 0 1 -3

I need to be 0.

I am a 0

Jeff Bivin -- LZHS

1 0 0 -1

0 1 0 2

0 0 1 -3

1 - 00 - 01 - 1

-4 – (-3)

31 RR

1 0 0 -1

0 1 0 2

0 0 1 -3

x = -1

y = 2

z = -3

Reading the Solution

Jeff Bivin -- LZHS

Writing the Solution

Jeff Bivin -- LZHS

x + y + z = -2 2x - 3y + z = -11

-x + 2y - z = 8