Assignment Solutions_a6s

# This is 1 4 3 5 9 5 9 a b 2 11 3 gaussian elimination

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Unformatted text preview: 3 + 4 = 1 and x1 − x2 − x4 = 1, so x1 = 1 + x2 + x4 = 1 + 1 = 2. 2 x1 x2 1 The solution is x3 = 1 . 0 x4 (e) [A|b] = [1 − 124]. This is row echelon form. The free variables are y = t and z = s, so x = 4 + y − 2z = 4 + t − 2s. In vector form the solution is x 4 + t − 2s 4 1 −2 y = = 0 + t 1 + s 0 . t z s 0 0 1 4. The question asks if there are scalars a and b such that 2 0 −1 0 −1 −11 = a −1 + b 4 ? This is −1 4 −3 5 9 5 9 a b 2 = −11 . −3 Gaussian elimination gives 0 −1 | 2 1 −4 | 11 1 −4 | 11 −1 4 | −11 → 0 −1 | 2 → 0 1 | −2 5 9 | −3 0 29 | −58 0 0| 0 There is a unique solution: b = −2, a = 3. Then the given vector is indeed a linear combination of the other two...
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## This note was uploaded on 02/13/2014 for the course MATH 2050 taught by Professor Dryuan during the Fall '12 term at Memorial University.

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