Linear Algebra Solutions 3

# Linear Algebra Solutions 3 - R 1 → R 1 + R 4 R 3 ↔ R 4...

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Case 1: a 2 - 16 6 = 0. i.e. a 6 = ± 4. Then R 3 1 a 2 - 16 R 3 R 1 R 1 - R 3 R 2 R 2 + 2 R 3 1 0 0 8 a +25 7( a +4) 0 1 0 10 a +54 7( a +4) 0 0 1 1 a +4 and we get the unique solution x = 8 a + 25 7( a + 4) , y = 10 a + 54 7( a + 4) , z = 1 a + 4 . Case 2: a = - 4. Then B = 1 0 1 8 7 0 1 - 2 10 7 0 0 0 - 8 , so our system is inconsistent. Case 3: a = 4. Then B = 1 0 1 8 7 0 1 - 2 10 7 0 0 0 0 . We read of that the system is consistent, with complete solution x = 8 7 - z, y = 10 7 + 2 z , where z is arbitrary. 12. We reduce the augmented array oF the system to reduced row–echelon Form: 1 0 1 0 1 0 1 0 1 1 1 1 1 1 0 0 0 1 1 0 R 3 R 3 + R 1 1 0 1 0 1 0 1 0 1 1 0 1 0 1 1 0 0 1 1 0 R 3 R 3 + R 2 1 0 1 0 1 0 1 0 1 1 0 0 0 0 0 0 0 1 1 0
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Unformatted text preview: R 1 → R 1 + R 4 R 3 ↔ R 4 1 1 1 1 1 1 1 1 . The last matrix is in reduced row–echelon Form and we read of the solution oF the corresponding homogeneous system: x 1 =-x 4-x 5 = x 4 + x 5 x 2 =-x 4-x 5 = x 4 + x 5 x 3 =-x 4 = x 4 , 6...
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## This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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