hw3solutions

hw3solutions - Math 215 HW#3 Solutions 1 Problem 1.6.6 Use...

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Math 215 HW #3 Solutions 1. Problem 1.6.6. Use the Gauss–Jordan method to invert A 1 = 1 0 0 1 1 1 0 0 1 , A 2 = 2 - 1 0 - 1 2 - 1 0 - 1 2 , A 3 = 0 0 1 0 1 1 1 1 1 . Solution: Start with the augmented matrix [ A 1 I ]: 1 0 0 1 0 0 1 1 1 0 1 0 0 0 1 0 0 1 . Then the only row on the left that doesn’t already look like the identity matrix is the second row; we just need subtract rows 1 and 3 from row 2, which gives: 1 0 0 1 0 0 0 1 0 - 1 1 - 1 0 0 1 0 0 1 . Hence, A - 1 1 = 1 0 0 - 1 1 - 1 0 0 1 To find A - 1 2 , start with the augmented matrix [ A 2 I ]: 2 - 1 0 1 0 0 - 1 2 - 1 0 1 0 0 - 1 2 0 0 1 . Replace the first row by half of itself and add half of the first row to the second: 1 - 1 2 0 1 2 0 0 0 3 2 - 1 1 2 1 0 0 - 1 2 0 0 1 . Next, add a third of the second row to the first, add 2/3 the second row to the third, and multiply the second row by 2/3: 1 0 - 1 3 2 3 1 3 0 0 1 - 2 3 1 3 2 3 0 0 0 4 3 1 3 2 3 1 . Finally, multiply the third row by 3 / 4, then add 1/3 of the result to row 1 and add 2/3 of the result to row 2: 1 0 0 3 4 1 2 1 4 0 1 0 1 2 1 1 2 0 0 1 1 4 1 2 3 4 . 1
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Thus, A - 1 2 = 3 4 1 2 1 4 1 2 1 1 2 1 4 1 2 3 4 . To find A - 1 3 , start with the augmented matrix [ A 3 I ]: 0 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 . First, switch rows 1 and 3: 1 1 1 0 0 1 0 1 1 0 1 0 0 0 1 1 0 0 . Now, subtract row 2 from row 1 and subtract row 3 from row 2: 1 0 0 0 - 1 1 0 1 0 - 1 1 0 0 0 1 1 0 0 . Thus, A - 1 3 = 0 - 1 1 - 1 1 0 1 0 0 . 2. Problem 1.6.8. Show that A = ± 1 1 3 3 ² has no inverse by solving Ax = 0, and by failing to solve ± 1 1 3 3 ²± a b c d ² = ± 1 0 0 1 ² . Solution: Note that (as discussed in class on Friday), ± 1 1 3 3 ²± 1 - 1 ² = ± 0 0 ² , so x = ± 1 - 1 ² is a solution of Ax = 0. The fact that this equation has such a solution implies that A is not invertible. To see this, note that if A were invertible, we could multiply both sides of the above equation by A - 1 , yielding x = A - 1 0 = 0. Since the given solution x is not zero, this is clearly impossible. Another proof that A is not invertible is as follows. If A were invertible, then there would exist A - 1 such that AA - 1 = I . Assuming A - 1 = ± a b c d ² , this means ± 1 1 3 3 ²± a b c d ² = ± 1 0 0 1 ² . 2
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This implies that a + c = 1 b + d = 0 3 a + 3 c = 0 3 b + 3 d = 1 . The third equation can be re-written as 3( a + c ) = 0 or, dividing both sides by 3, as a + c = 0. But this directly contradicts the first equation, meaning that there is no solution to this system; equivalently, A is not invertible. 3. Problem 1.6.14. If
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hw3solutions - Math 215 HW#3 Solutions 1 Problem 1.6.6 Use...

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