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solution2 - p 21 Exercise 2 Let Homework assignment 2 1 1 1...

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Let A = 1 - 1 1 2 0 1 3 0 1 , B = 2 - 2 1 3 4 4 Verify directly that A ( AB ) = A 2 B A 2 = 2 - 1 1 5 - 2 3 6 - 3 4 , A 2 B = 7 - 3 20 - 4 25 - 5 AB = 5 - 1 8 0 10 - 2 , A ( AB ) = 7 - 3 20 - 4 25 - 5 Find two different 2 × 2 matrices A such that A 2 = 0 but A 6 = 0 . 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 0 but their squares are 0 . 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
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Let us first row-reduce A into the identity matrix: A - 2 · I + II + 3 1 - 1 1 0 2 - 1 3 0 1 - 3 · I + III + 3 1 - 1 1 0 2 - 1 0 3 - 2 1 2 · II + 3 1 - 1 1 0 1 - 1 2 0 3 - 2 - 3 · II + III + 3 1 - 1 1 0 1 - 1 2 0 0 - 1 2 - 2 · III + 3 1 - 1 1 0 1 - 1 2 0 0 1 - 1 2 · III + II + 3 1 - 1 1 0 1 0 0 0 1 - 1 · III + I + 3 1 - 1 0 0 1 0 0 0 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 0 1 Analogously we obtain all the other matrices, the last one is E 8 = 1 1 0 0 1 0 0 0 1 Let A and B be 2 × 2 matrices such that AB = I . Prove that BA = I . First proof: Let A = a b c d , B = x 1 x 2 x 3 x 4 . Consider x i as unknowns, and a, b, c, d as coefficients. Since AB = I , the un- knowns x 1 , x 3 satisfy the following 2 linear equations ax 1 + bx 3 = 1 cx 1 + dx 3 = 0 , while the unknowns x 2 , x 4 satify the equations ax 2 + bx 4 = 0 cx 2 + dx 4 = 1 .
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