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Transcript
Linear Algebra Problem Set 2 Solutions 38 14 25 / 54 7 / 27 1 1. PQ , ( PQ ) 32 / 27 19 / 27 64 25 4 / 27 1 / 27 1 / 2 3 / 2 25 / 54 7 / 27 1 1 1 , , P 1 Q Q P 1 32 / 27 19 / 27 4 7 / 27 5 / 27 2. Assume that there is an inverse and then try and find it. If there is an inverse then: 0 B X Y I 0 C D W Z 0 I If we can find X, Y, W and Z then we have found the inverse. Do the multiplication on the left and set it equal to the right. 0 X BW 0Y BZ I 0 CX DW CY DZ 0 I 0 X BW I BW I W B 1 0Y BZ 0 BZ 0 Z 0 CX DW 0 CX DB 1 0 CX DB 1 X C 1 DB 1 CY DZ I CY D0 I CY I Y C 1 So the inverse is: 1 B C 1 DB 1 C 1 D B 1 0 C and B must be invertible. 0 C 3 4 x 2 5 / 31 4 / 31 3. , A 1 , 4 5 y 1 4 / 31 3 / 31 6 / 31 A 1b x x 6 / 11, y 11 / 31 11 / 31 4. Check by showing AA1 I . Details are below. 5. a) Let A be a square matrix with an ith row of zeros. Let B A1 . Then AB I . The ith row of AB ith row of A * B = [0 0 0 ….. 0]. Since AB I then ( AB) ii 1 , but ( AB) ii 0 . Thus a matrix with a row of all zeros is not invertible. b) A similar argument is used to show if a matrix has a column of zeros it is not invertible. Use the fact that BA I and reach a contradiction. 6. a) False: The zero matrix is a counterexample. Any non-invertible matrix will not be able to be written as the product elementary matrices. 3 0 1 0 b) False: Let E1 and E 2 are both elementary matrices, but 0 1 3 1 3 0 E1 E 2 is not an elementary matrix. 3 1 c) True: Multiplying a row of matrix A by a constant and adding it to another row is an elementary row operation and can be written as E1 A . If A is invertible then this product is invertible because an elementary matrix is invertible and the product of invertible matrices is invertible. d) True: AB 0 A 1 AB A 1 0 IB 0 B 0 7. 1 2 1 b1 4 5 2 b2 4 7 4 b3 After Gauss Elimination the matrix becomes: 1 0 0 3b1 1 / 2b2 3 / 2b3 0 1 0 4b1 b3 0 0 1 4b1 1 / 2b2 3 / 2b3 The system is consistent for all b’s so no restrictions. x1 3b1 1 / 2b2 3 / 2b3 x 2 4b1 b3 x3 4b1 1 / 2b2 3 / 2b3