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Systems of Equations Gaussian Elimination & Row Reduced Echelon Form by Jeffrey Bivin Lake Zurich High School Last Updated: October.

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Presentation on theme: "Systems of Equations Gaussian Elimination & Row Reduced Echelon Form by Jeffrey Bivin Lake Zurich High School Last Updated: October."— Presentation transcript:

1 Systems of Equations Gaussian Elimination & Row Reduced Echelon Form by Jeffrey Bivin Lake Zurich High School jeff.bivin@lz95.org Last Updated: October 17, 2005

2 Example 1 Jeff Bivin -- LZHS

3 x + y + z = 6 4x – 8y + 4z = 12 2x – 3y + 4z = 3 1116 4-8412 2-343 Jeff Bivin -- LZHS

4 1116 4-8412 2-343 I am a 1. Jeff Bivin -- LZHS

5 1116 4-8412 2-343 I need to be 0. Jeff Bivin -- LZHS 1116 0-120 0-52-9 4 - 4(1) -8 - 4(1) 4 - 4(1) 12 - 4(6) 2 - 2(1) -3 - 2(1) 4 - 2(1) 3 - 2(6)

6 1116 0-120 0-52-9 I need to be 1 Jeff Bivin -- LZHS 1116 0101 0-52-9

7 1116 0101 0-52-9 I need to be 0. Jeff Bivin -- LZHS 1015 0101 002-4 1 - 0 1 - 1 1 - 0 6 - 1 0 + 5(0) -5 + 5(1) 2 + 5(0) -9 + 5(1)

8 1015 0101 002-4 I need to be 1 Jeff Bivin -- LZHS 1015 0101 001-2

9 1015 0101 001 I need to be 0. I am a 0 Jeff Bivin -- LZHS 1007 0101 001-2 1 - 0 0 - 0 1 - 1 5 – (-2)

10 1007 0101 001-2 x = 7 y = 1 z = -2 Reading the Solution Jeff Bivin -- LZHS

11 Writing the Solution x + y + z = 6 4x – 8y + 4z = 12 2x – 3y + 4z = 3 Jeff Bivin -- LZHS

12 Example 2 Jeff Bivin -- LZHS

13 x + y + z = -2 2x - 3y + z = -11 -x + 2y - z = 8 111-2 2-31-11 2 8 Jeff Bivin -- LZHS

14 111-2 2-31-11 2 8 I am a 1. Jeff Bivin -- LZHS

15 111-2 2-31-11 2 8 I need to be 0. Jeff Bivin -- LZHS 111-2 0-5-7 0306 2 - 2(1) -3 - 2(1) 1 - 2(1) -11 - 2(-2) -1 + 1 2 + 1 -1 + 1 8 + (-2)

16 111-2 0-5-7 0306 I would prefer to make the 3 a one in row three rather than the -5 in row 2. Why? Jeff Bivin -- LZHS 111-2 0306 0-5-7 To avoid fractions! We will switch Row 2 and Row 3

17 111-2 0306 0-5-7 I need to be 1 Jeff Bivin -- LZHS 111-2 0102 0-5-7

18 111-2 0102 0-5-7 I need to be 0. Jeff Bivin -- LZHS 101-4 0102 003 1 - 0 1 - 1 1 - 0 -2 - 2 0 + 5(0) -5 + 5(1) -1 + 5(0) -7 + 5(2)

19 101-4 0102 003 I need to be 1 Jeff Bivin -- LZHS 101-4 0102 001-3

20 101-4 0102 001-3 I need to be 0. I am a 0 Jeff Bivin -- LZHS 100 0102 001-3 1 - 0 0 - 0 1 - 1 -4 – (-3)

21 100 0102 001-3 x = -1 y = 2 z = -3 Reading the Solution Jeff Bivin -- LZHS

22 Writing the Solution Jeff Bivin -- LZHS x + y + z = -2 2x - 3y + z = -11 -x + 2y - z = 8


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