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Unformatted text preview: John Q. Student MATH 211 (Siefken): Homework #2 Due May 26, 2011 Question #1 (a) Find the inverse of A = 3 1 2 6 3 5 1 1 3 by GaussJordan elimination. To find A 1 we apply GaussJordan elimination to [ A  I ]. 3 1 2 1 0 0 6 3 5 0 1 0 1 1 3 0 0 1 R 2 7 R 2 2 R 1 3 1 2 1 0 0 0 1 1 2 1 0 1 1 3 0 1 R 3 7 R 3 1 / 3 R 1 3 1 2 1 0 0 1 1 2 1 0 0 2 / 3 7 / 3 1 / 3 0 1 R 3 7 R 3 2 / 3 R 2 3 1 2 1 0 1 1 2 1 0 0 5 / 3 1 2 / 3 1 R 2 7 R 2 3 / 5 R 3 3 1 2 1 0 1 13 / 5 7 / 5 3 / 5 0 0 5 / 3 1 2 / 3 1 R 1 7 R 1 6 / 5 R 3 3 1 1 / 5 4 / 5 6 / 5 0 1 13 / 5 7 / 5 3 / 5 0 0 5 / 3 1 2 / 3 1 R 1 7 R 1 R 2 3 0 12 / 5 3 / 5 3 / 5 0 1 13 / 5 7 / 5 3 / 5 0 0 5 / 3 1 2 / 3 1 R 1 7 R 1 / 3 1 0 4 / 5 1 / 5 1 / 5 0 1 13 / 5 7 / 5 3 / 5 0 0 5 / 3 1 2 / 3 1 R 3 7 3 / 5 R 3 1 0 0 4 / 5 1 / 5 1 / 5 0 1 0 13 / 5 7 / 5 3 / 5 0 0 1 3 / 5 2 / 5 3 / 5 and so we see A 1 = 1 5 4 1 1 13 7 3 3 2 3 . (b) Solve the following system of equations 3 x 1 y 2 z = 1 6 x 3 y 5 z = 2 x + y +3 z = 3 . We notice that A is th coefficient matrix for the system of equations given. That is, the system of equations is equivalent to the matrix equation A x =  1 2 3 . Since we have already computed A 1 , multiplying both sides by A 1 gives x y z = 1 5 4 1 1 13 7 3 3 2 3  1 2 3 =  9 / 5 18 / 5 2 / 5 . Question #2 For each of the following statements, if they are true, prove them , otherwise provide a specific counter example . (a) If A and B are both 2 2 matrices which are not diagonal (and so neither is the zero matrix), then AB = BA implies A = B . Question #2 continued on next page... Page 1 of 6 John Q. Student MATH 211 (Siefken): Homework #2 Due May 26, 2011 Consider A = 3 1 1 3 and B = 3 1 1 3 . We have AB = BA = 8 0 0 8 , but A 6 = B ....
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 Summer '10
 SOSPEDRAALFONSO
 Linear Algebra, Algebra, GaussJordan Elimination

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