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Unformatted text preview: Homework assignment 2 p 21 Exercise 2. Let A = 1 1 1 2 1 3 1 , B = 2 2 1 3 4 4 Verify directly that A ( AB ) = A 2 B Solution: A 2 = 2 1 1 5 2 3 6 3 4 , A 2 B = 7 3 20 4 25 5 AB = 5 1 8 10 2 , A ( AB ) = 7 3 20 4 25 5 Exercise 3. Find two different 2 2 matrices A such that A 2 = 0 but A 6 = 0 . Solution: Let A = a b c d be a 2 2 matrix. Then A 2 = a 2 + bc ( a + d ) b ( a + d ) c d 2 + bc . Now find a,b,c,d such that a 2 + bc = ( a + d ) b = ( a + d ) c = d 2 + bc = 0 . Note that if a + d = 0 and ad bc = 0 , then all these equations are satisfied. For instance, put a = d = 1 ,b = c = 1 or put a = d = 0 ,b = 0 ,c = 1 . Then the matrices 1 1 1 1 , 0 1 0 0 are not but their squares are . Exercise 4. For the matrix A of Exercise 2 , find elementary matrices E 1 ,E 2 ,...,E k such that E k E 2 E 1 A = I 1 Solution: Let us first rowreduce A into the identity matrix: A 2 I + II + 3 1 1 1 2 1 3 1  3 I + III + 3 1 1 1 2 1 3 2 1 2 II + 3 1 1 1 1 1 2 3 2  3 II + III + 3 1 1 1 1 1 2 1 2  2 III + 3 1 1 1 1 1 2 1  1 2 III + II + 3 1 1 1 1 1  1 III + I + 3 1 1 0 1 1 1 II + I + 3 1 0 0 0 1 0 0 0 1 Applying the transformation on top of each arrow to the identity matrix, we obtain the elementary transformations we want. For example, E 1 is obtained by applying 2 I + II to the identity matrix, therefore: E 1 = 1 0 0 2 1 0 0 1 Analogously we obtain all the other matrices, the last one is E 8 = 1 1 0 0 1 0 0 0 1 Exercise 7. Let A and B be 2 2 matrices such that AB = I . Prove that BA = I ....
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This note was uploaded on 11/14/2010 for the course PHYCS 498 taught by Professor Aa during the Spring '10 term at University of Illinois, Urbana Champaign.
 Spring '10
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