Linear Algebra Solutions 49

# Linear Algebra Solutions 49 - R1 → R1 − R2 R2 → R2...

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Unformatted text preview: R1 → R1 − R2 R2 → R2 − R3 R3 → R3 − R4 = C2 → C 2 + C 1 = C3 → C 3 + C 2 c = c 0 0 0 b 2b + c −b − c −b 0 0 c −c a 2a a+b a+b+c =c 2b + c −b − c −2b − c 0 c 0 2a a + b 2a + 2b + c = c2 = c2 (2b + c) 17. Let ∆ = c −c 0 0 b b + c −b − c −b 0 0 c −c a a a+b a+b+c 1 −1 2a 2a + 2b + c 2b + c −b − c −b 0 c −c 2a a+b a+b+c 2b + c −2b − c 2a 2a + 2b + c = c2 (2b + c)(4a + 2b + c). 1 + u1 u1 u1 u1 u2 1 + u2 u2 u2 u3 u3 1 + u3 u3 u4 u4 u4 1 + u4 . Then using the operation R1 → R1 + R2 + R3 + R4 we have ∆= t t t t u2 1 + u 2 u2 u2 u3 u3 1 + u3 u3 u4 u4 u4 1 + u4 (where t = 1 + u1 + u2 + u3 + u4 ) 1 1 1 1 u2 1 + u 2 u2 u2 = (1 + u1 + u2 + u3 + u4 ) u3 u3 1 + u3 u3 u4 u4 u4 1 + u4 The last determinant equals C2 → C 2 − C 1 C3 → C 3 − C 1 C4 → C 4 − C 1 1 u2 u3 u4 52 0 1 0 0 0 0 1 0 0 0 0 1 = 1. ...
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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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