Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# 4 the equations are x c 2y d 1 and 3x c 6y d 0 no

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Unformatted text preview: roof of .AB/c D A.B c / used the column rule for matrix multiplication—this rule is clearly linear, column by column. Even for nonlinear transformations, A.B.c // would be the “composition” of A with B (applying B then A). This composition A ı B is just AB for matrices. One of many uses for the associative law: The left-inverse B = right-inverse C from B D B.AC / D .BA/C D C . Problem Set 2.5, page 89 1A 1 D  0 1 3 1 4 0  and B 1 D  1 2 1 0 1 2  and C 2 A simple row exchange has P 2 D I so P P 1 1 1 D  7 5 D P . Here P = “transpose” of P , coming in Section 2:7. 1  4 . 3 " D # 001 1 0 0 . Always 010 Solutions to Exercises 18         1 x :5 t :2 5 2 1 3 D and D so A D . This question solved y :2 z :1 2 1 10 1 AA D I column by column, the main idea of Gauss-Jordan elimination. 4 The equations are x C 2y D 1 and 3x C 6y D 0. No solution because 3 times equation 1 gives 3x C 6y D 3.   a for any a. And also U . 1 1 5 An upper triangular U with U D I is U D 0 2 6 (a) Multiply AB D AC by A  as B x x C has the form 1 to ﬁnd B D C (since A is invertible) (b) As long   y 11 , we have AB D AC for A D . y 11 7 (a) In Ax D .1; 0; 0/, equation 1 C equation 2 equation 3 is 0 D 1 (b) Right (c) Row 3 becomes a row of zeros—no third pivot. sides must satisfy b1 C b2 D b3 8 (a) The vector x D .1; 1; 1/ solves Ax D 0 (b) After elimination, columns 1 and 2 end in zeros. Then so does column 3 D column 1 C 2: no third pivot. 1 9 If you exchange rows 1 and 2 of A to reach B , you exchange columns 1 and 2 of A 1 to reach B 2 1 . In matrix notation, B D PA has B 3 2 0 0 0 1=5 0 1=4 0 7 60 6 and B 1 D 4 10 A 1 D 4 0 1=3 0 05 1=2 0 0 0 block of B ). DA 3 4 0 0 1 1 P 2 3 0 0 1 D A P for this P . 3 0 0 0 07 (invert each 6 55 7 6   10 A then certainly A C B = zero matrix is not invertible. (b) A D 00   00 are both singular but A C B D I is invertible. and B D 01 11 (a) If B D 12 Multiply C D AB on the left by A 13 M 1 CM 14 B 1 DC A. 1 1 DA 1  B 1 1 1 0 1 A  1 1 and on the right by C 1 : Then A 1 D BC so multiply on the left by C and the right by A W B 1 DA 1   10 : subtract column 2 of A 11 1 1 1 . D from column 1. 1 .  d b ad b c 0 The inverse of each matrix is ab D . 16 the other divided by ad b c cd c a 0 ad b c # # " " #" #" 1 1 1 1 1 1 1 1 11 D E. D 17 E32 E31 E21 D 0 11 1 11 1 1 " # 1 Reverse the order and change 1 to C1 to get inverses E211 E311 E321 D 1 1 D 111 L D E 1 . Notice the 1’s unchanged by multiplying in this order. 15 If A has a column of zeros, so does BA. Then BA D I is impossible. There is no A     18 A2 B D I can also be written as A.AB/ D I . Therefore A 1 is AB . Solutions to Exercises 19 19 The .1; 1/ entry requires 4a bD aD 1 5 2 . 6 and a D 2 . 5 3b D 1; the .1; 2/ entry requires 2b a D 0. Then For the 5 by 5 case 5a 4 b D 1 and 2b D a give b D 1 and 6 20 A  ones.4; 1/ is the zero vector so A cannot be invertible. 21 Six of the sixteen 0 1 matrices are invertible, including all four with thre...
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## This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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