Assignment Solutions_a6s

# Or from the last row we can see that there is no

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Unformatted text preview: − 2 7 14 −7 −4 1 7 −6 0 6 −42 3 −18 1 0 0 0 1 7 6 0 0 −42 0 0 | 2 | −5 | −2 | −7 | 2 | 5 . | −7 | − 13 2 The rank of A is 3 while the rank of the augmented matrix [A|b] is 4, so the system has no solution. (OR: From the last row, we can see that there is no solution.) (c) 2 2 2 −8 | 1 2 2 2 −8 | 1 0 | 4 → 0 2 2 16 | 2. [A|b] = 4 6 6 6 6 10 −4 | 2 0 0 4 20 | −1 This is upper triangular form. We can ﬁnd the solutions by back substitution. Variable x4 = t is free and 4x3 +20x4 = −1, so x3 = − 1 −5x4 = − 1 −5t. Then x2 = 1−x3 −8x4 = 4 4 5 4 − 3t and x1 = 1 2 − x2 − x3 + 4x4 = − 1 + 12t. The solution 2 1 1 −2 12 − 2 + 12t −3 5 − 3t 5 4 x = 41 − − 5t = − 1 + t −5 4 4 t 0 1 is . (d) 2 −3 −1 1 [A|b] = 0 2 3 0 1 −1 0 −1 → 0 2 0 3 0 −1 0 1 → 0 0 0 0 1 −1 0 1 → 0 0 0 0 1 −1 0 1 → 0 0 0 0 | 5 1 −1 0 −1 | 1 | −1 4 −1 | 5 → 2 −3 0 | 1 2 −1 −...
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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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