Systems of Equations Gaussian Elimination & Row Reduced Echelon Form by Jeffrey Bivin Lake Zurich High School Last Updated: October 17, 2005

Example 1 Jeff Bivin -- LZHS

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

I am a 1. Jeff Bivin -- LZHS

I need to be 0. Jeff Bivin -- LZHS (1) (1) 4 - 4(1) (6) 2 - 2(1) (1) 4 - 2(1) 3 - 2(6)

I need to be 1 Jeff Bivin -- LZHS

I need to be 0. Jeff Bivin -- LZHS (0) (1) 2 + 5(0) (1)

I need to be 1 Jeff Bivin -- LZHS

I need to be 0. I am a 0 Jeff Bivin -- LZHS – (-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 = Jeff Bivin -- LZHS

I am a 1. Jeff Bivin -- LZHS

I need to be 0. Jeff Bivin -- LZHS (1) (1) 1 - 2(1) (-2) (-2)

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

I need to be 1 Jeff Bivin -- LZHS

I need to be 0. Jeff Bivin -- LZHS (0) (1) (0) (2)

I need to be 1 Jeff Bivin -- LZHS

I need to be 0. I am a 0 Jeff Bivin -- LZHS – (-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