# hw4_solutions - CS 257 Numerical Methods - Homework 4...

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CS 257 Numerical Methods - Homework 4 October 5, 2006 1. [1pt] Section 7.1 #1 Solution: A = 1 4 α 2 - 1 2 α α 3 1 b = 6 3 5 det( A ) = 9( α - 1)( α + 1) α = 0: The determinant of A is non-zero, hence there exists a unique solution. α = 1: The determent of A is zero. To see if there are inﬁnitely many solutions or no solutions, we use Gaussian Elimination to create an equivalent system. ˜ A = 1 0 1 0 1 0 0 0 0 ˜ b = 2 1 0 We eﬀectively have two equations and three unknowns. Note that x 2 = 1. We are free to pick a value for x 3 and ﬁnd a value for x 1 . There are inﬁnitely many solutions. α = - 1: The determinant of A is again zero and we create an equivalent system. ˜ A = 1 0 - 1 0 1 0 0 0 0 ˜ b = 0 0 1 The last equation implies 0 = 1, which is false. This system has no solution. If we replace the RHS by b = [0 0 0] T , then setting all three variables to 0 always solves the system. α = 0: The determinant of A is non-zero. The solution we found is unique. α = 1: Because the determinant is zero AND there exists one solution, we know there are inﬁnitely many solutions. α = - 1: By the same reasoning as the above point, there are inﬁnitely many solutions. 2. [1pt] Section 7.1 #4 Solution: I will use rounding in my solution. I will show the augmented matrix after each elimination. All binary operations are performed only to the given accuracy. Four digits of precision: ± 0 . 1036 0 . 2122 0 . 7381 0 . 2081 0 . 4247 0 . 9327 ² ± 0 . 1036 0 . 2122 0 . 7381 0 - 0 . 001600 - 0 . 5503 ² In the back substitution, we ﬁnd x 2 = - 0 . 5503 0 . 001600 = 343 . 9 x 1 = 0 . 7381 - 0 . 2122 * x 2 0 . 1036 = 0 . 7381 - 72 . 98 0 . 1036 = - 72 . 24 0 . 1036 = - 697 . 3 1

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Eight digits of precision: ± 0 . 1036 0 . 2122 0 . 7381 0 . 2081 0 . 4247 0 . 9327 ² ± 0 . 1036 0 . 2122 0 . 7381 0 - 0 . 00154345 - 0 . 54991210 ² In the back substitution, we ﬁnd x 2 = - 0 . 54991210 - 0 . 00154345 = 356 . 28760 x 1 = 0 . 7381 - 0 . 2122 * x 2 0 . 1036 = 0 . 7381 - 75 . 604229 0 . 1036 = - 74 . 866129 0 . 1036 = - 722 . 64603 3. [1pt] Section 7.1 #7a Solution: 3 4 3 10 1 5 - 1 7 6 3 7 15 3 4 3 10 0 3 . 667 - 2 3 . 667 0 - 5 1 - 5 3 4 3 10 0 3 . 667 - 2 3 . 667 0 0 - 1 . 728 . 001788 Performing the back substitution x 3 = . 001788 - 1
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## This note was uploaded on 09/15/2008 for the course CS 257 taught by Professor Thomaskerkhoven during the Fall '05 term at University of Illinois at Urbana–Champaign.

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hw4_solutions - CS 257 Numerical Methods - Homework 4...

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