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Unformatted text preview: Math 215 HW #8 Solutions 1. Problem 4.2.4. By applying row operations to produce an upper triangular U , compute det 1 2 2 0 2 3 4 1 1 2 2 2 5 3 and det 2 1 1 2 1 1 2 1 1 2 . Answer: Focusing on the first matrix, we can subtract twice row 1 from row 2 and add row 1 to row 3 to get 1 2 2 0 1 1 2 2 2 5 3 . Next, add twice row 2 to row 4: 1 2 2 0 1 1 2 2 5 5 . Finally, add 5 / 2 times row 3 to row 4: 1 2 2 1 1 2 2 10 . Since none of the above row operations changed the determinant and since the determinant of a triangular matrix is the product of the diagonal entries, we see that det 1 2 2 0 2 3 4 1 1 2 2 2 5 3 = (1)( 1)( 2)(10) = 20 . Turning to the second matrix, we can first add half of row 1 to row 2: 2 1 0 3 / 2 1 1 2 1 1 2 . Next, add 2 / 3 of row 2 to row 3: 2 1 0 3 / 2 1 4 / 3 1 1 2 . 1 Finally, add 3 / 4 of row 3 to row 4: 2 1 0 3 / 2 1 4 / 3 1 5 / 4 . Therefore, since the row operations didnt change the determinant and since the determinant of a triangular matrix is the product of the diagonal entries, det 2 1 1 2 1 1 2 1 1 2 = (2)(3 / 2)(4 / 3)(5 / 4) = 5! 4! = 5 . Note : This second matrix is the same one that came to our attention in Section 1.7 and HW #3, Problem 9. 2. Problem 4.2.6. For each n , how many exchanges will put (row n , row n 1, ... , row 1) into the normal order (row 1, ... , row n 1, row n )? Find det P for the n by n permutation with 1s on the reverse diagonal. Answer: Suppose n = 2 m is even. Then the following sequence of numbers gives the original ordering of the rows: 2 m, 2 m 1 ,...,m + 1 ,m,..., 2 , 1 . Exchanging 2 m and 1, and then 2 m 1 and 2, ... , and then m + 1 and m yields the correct ordering of rows: 1 , 2 ,...,m,m + 1 ,..., 2 m 1 , 2 m. Clearly, we performed m = n/ 2 row exchanges in the above procedure. Thus, for even values of n , we need to perform n/ 2 row exchanges. On the other hand, suppose n = 2 m 1 is odd. Then the original ordering of the rows is 2 m 1 , 2 m 2 ,...,m + 1 ,m,m 1 ,..., 2 , 1 . We exchange 2 m 1 and 1, and then 2 m 2 and 2, ... , and then m + 1 and m 1. Since m is already in the correct spot, this gives the correct ordering of rows 1 , 2 ,...,m 1 ,m,m + 1 ,..., 2 m 2 , 2 m 1 . Clearly, we performed m 1 = n 1 2 row exchanges. Thus, for odd values of n , we need to perform n 1 2 row exchanges. If P is the permutation matrix with 1s on the reverse diagonal, then the rows of P are simply the rows of the identity matrix in precisely the reverse order. Thus, the above reasoning tells us how many row exchanges will transform P into I . Since the determinant of the identity matrix is 1 and since performing a row exchange reverses the sign of the determinant, we...
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 Spring '09
 Math

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