plugin-hw04 - Assignment 4 Math 417 Winter 2009 Due...

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Assignment 4 Math 417 — Winter 2009 Due February 6 § 2.3#6. a b c d d - b - c a = ad - bc - ab + ba cd - dc - bc + ad = ad - bc 0 0 ad - bc . § 2.3#14. As always, the product of two matrices is defined if the number of columns of the first equals the number of rows of the second. So we can multiply either of D and E (which have only 1 column) by either of B and E (which have only 1 row); we can multiply A (which has 2 columns) by A (which has 2 rows); and we can multiply either of B and C (which have 3 columns) by either of C and D (which have 3 rows). The products are: DB = 1 2 3 1 2 3 1 2 3 , DE = 5 5 5 , EB = 5 10 15 , EE = [25] , AA = 2 2 2 2 , and BC = 14 8 2 , BD = [6] , CC = - 2 - 2 - 2 4 1 - 2 10 4 - 2 , CD = 0 3 6 . § 2.3#18. For a matrix X = x 1 x 2 x 3 x 4 to commute with the given matrix A means that the two products AX = x 1 + 2 x 3 x 2 + 2 x 4 x 3 x 4 and XA = x 1 2 x 1 + x 2 x 3 2 x 3 + x 4 must be equal. This gives us four linear equations x 1 + 2 x 3 = x 1 , x 2 + 2 x 4 = 2 x 1 + x 2 , x 3 = x 3 , x 4 = 2 x 3 + x 4 . 1
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Reducing this system we arrive at the two equations x 1 = x 4 , x 3 = 0, and the two variables x 2 , x 4 are free. Assigning x 2 = s and x 4 = t , we see that the matrices which commute with A are exactly those of the form t s 0 t . § 2.3#26. Proceeding just as in the preceding problem, we have that X = x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 commutes with the given A if and only if the two products AX = 2 x 1 2 x 2 2 x 3 2 x 4 2 x 5 2 x 6 3 x 7 3 x 8 3 x 9 and XA = 2 x 1 2 x 2 3 x 3 2 x 4 2 x 5 3 x 6 2 x 7 2 x 8 3 x 9 are equal. The entries in the upper left 2 × 2 blocks are automatically equal, no matter what the entries of X are, and the same goes for the bottom right entry. Only the first two entries in the last row and column impose any requirements on X , and they say that x 3 = x 6 = x 7 = x 8 = 0. Thus, the matrices that commute with A are those of the form * * 0 * * 0 0 0 * . § 2.4#10. Row reduce the given matrix augmented by the identity matrix: 1 1 1 1 0 0 1 2 3 0 1 0 1 3 6 0 0 1 = 1 1 1 1 0 0 0 1 2 - 1 1 0 0 2 5 - 1 0 1 R 2 R 2 - R 1 R 3 R 3 - R 1 = 1 0 - 1 2 - 1 0 0 1 2 - 1 1 0 0 0 1 1 - 2 1 R 1 R 1 - R 2 R 3 R 3
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