Finite Lab 3 Solutions

# Finite Lab 3 Solutions - Math 120 Spring 2008 Brodnick Lab...

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Unformatted text preview: Math 120 Spring 2008 Brodnick Lab #3: 2.2, 2.3, 3.1, & 3.2 Solutions 1. 6 - 1 4 3 a - 2c 6 6 R + 1R 0 3a 2 1 4 22 a - 2c 22 R1 - 4 R2 132 19a - 2c 0 1 0 0 1 0 22 1 - 54a - 36c 132 R1 1 19a - 2c 22 R2 9 3 - 22 a - 11 c 19 1 a - 11 c 22 d 23 = 19 22 1 a - 11 c n + d + q = 21 2. 0.05n + 0.1d + 0.25q = 3.35 n + 1 = d n - d = -1 - 4 2 A - 3B T = 10 1 1 1 21 1 RREF .05 .1 .25 3.35 = 0 1 -1 0 - 1 0 9 20 = 6c 13 5 - 8 - 6c 0 1 0 0 0 1 5 6 10 quarters 3. 14 - 24 - -8 -3 4. 4 3 2 a -2 -3 c 1 1 -1 8 - 3a - 2 - 4 + 4a - 10 - 3a + 6 4a - 14 4 = = 6 - 3c + 1 - 3 + 4c + 5 - 3c + 7 4c + 2 5 2 0 1 -3 0 -2 - 4 x 4 - 1 y = 3 1 z - 8 5. 2x - 3y - 4z = 4 -z=3 x - 2 y + z = -8 ...
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## This note was uploaded on 04/07/2008 for the course MAT 120 taught by Professor Brodnick during the Spring '08 term at Illinois State.

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