FinalTestReviewSpring2011

# 5 a use guassian elimination represent the linear

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Unformatted text preview: (a) Use Guassian elimination represent the linear system A￿ = ￿ in an upper triangular to xb 1 −1 −1 x 2 2 −2 , ￿ = y , and ￿ = 0 . b form U ￿ = ￿, where A = 2 xc x 3 3 3 z 6 (12 pts) Solution: [A : ￿ ] = b 1 −1 −1 2 1 −1 −1 2 4 0 −4 2 −2 0 −2￿1 → 0 r 2 −3￿1 r 3 3 3 6 3 3 36 1 −1 −1 2 1 −1 −1 2 4 0 −4 4 0 −4 → 0 → 0 3 − 2 ￿2 0 6 6 0 0 0 6 6 r 1 −1 −1 2 4 0 and ￿ = −4 . So, in U ￿ = ￿ we have U = 0 xc c 0 0 6 6 (b) Use part (a) and back substitution to solve the above linear system A￿ = ￿ . xb (10 pts) Solution: U ￿ = ￿ gives us the equations: xc 6z = 6, so z = 1; 4y = −4, so y = −1; x − y − z = 2, so x = 2 + y + z = 2 − 1 + 1 = 2. x 2 Answer: ￿ = y = −1 x z 1 (c) Use part (a) to ﬁnd a lower triangular matrix L for which A = LU . (11 pts) Solution: The elimination matrices were, consecutively, E21 Their respective inverses are − E211 100 100 1 00 1 0 . = −2 1 0 , E31 = 0 1 0 , and E32 = 0 3 001 −3 0 1 0 −2 1 100 100 100 −...
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