hmw3_sol_sp08

# hmw3_sol_sp08 - i ECE220 Homework Assignment#3 Solutions...

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i ECE220 Homework Assignment #3 Solutions Problem 6.21. Let a and b be real-valued constants. Let 2 3 a - 1 - 2 2 - 1 3 4 x 1 x 2 x 3 = 2 b 1 Determine, if possible, values for a and b that will make the system have no solution (i.e., inconsistent), one solution exactly inﬁnite solutions Solution The augmented matrix is 2 3 a 2 - 1 - 2 2 b - 1 3 4 1 . Performing Gaussian elimination: Row 1 Row 1 + Row 2 1 1 a + 2 b + 2 - 1 - 2 2 b - 1 3 4 1 Row 2 Row 2 - Row 3 1 1 a + 2 b + 2 0 - 5 - 2 b - 1 - 1 3 4 1 Row 3 Row 3 + Row 1 1 1 a + 2 b + 2 0 - 5 - 2 b - 1 0 4 a + 6 b + 3 Row 1 Row 1 + 1/5 * Row 2

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ii 1 0 5 a +8 5 6 b +9 5 0 - 5 - 2 b - 1 0 4 a + 6 b + 3 Row 2 -1/5 * Row 2 1 0 5 a +8 5 6 b +9 5 0 1 2 / 5 1 - b 5 0 4 a + 6 b + 3 Row 3 Row 3 - 4 * Row 2 1 0 5 a +8 5 6 b +9 5 0 1 2 / 5 1 - b 5 0 0 5 a +22 5 9 b +11 5 This is an upper triangle matrix with the last row implying 5 a + 22 5 x 3 = 9 b + 11 5 (0.0) The system of equations have No solution if 5 a +22 5 = 0 and 9 b +11 5 6 = 0 . a = - 22 / 5 ,b 6 = - 11 / 9 . Unique solution if 5 a +22 5 6 = 0 a 6 = - 22 / 5 . Inﬁnite solutions if 5 a +22 5 = 0 and 9 b +11 5 = 0 . a = - 22 / 5 ,b = - 11 / 9 . ————————————————————————————– Problem 6.25. Solve, if possible, the following system, using Gaussian elimination. 2 - 5 4 1 - 2 1 1 - 4 6 x 1 x 2 x 3 = - 3 5 - 1 Solution m = 3 ,n = 3 . Since the number of equations and variables are the same, the system is neither overdetermined nor underdetermined. The augmented matrix is 2 - 5 4 - 3 1 - 2 1 5 1 - 4 6 - 1
Performing Gaussian elimination: Row 1 Row 1 - Row 2 1 - 3 3 - 8 1 - 2 1 5 1 - 4 6 - 1 Row 2 Row 2 - Row 3 1 - 3 3 - 8 0 2 - 5 6 1 - 4 6 - 1 Row 3 Row 3 - Row 1 1 - 3 3 - 8 0 2 - 5 6 0 - 1 3 7 Row 1 Row 1 - 3 * Row 3 1 0 - 6 - 29 0 2 - 5 6 0 - 1 3 7 Row 2 1/2 * Row 2 1 0 - 6 - 29 0 1 - 5 / 2 3 0 - 1 3 7 Row 3 Row 3 + Row 2 1 0

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hmw3_sol_sp08 - i ECE220 Homework Assignment#3 Solutions...

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