Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# Then a 1 d de12 e21 d 0 12 6 2 1 2 2 1 1 0 0 2 1 0 1

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Unformatted text preview: e 1’s.      1310 1 3 1 0 1 0 7 3 22 ! ! D I A1; 2701 0 1 2 1 0 1 2 1         1410 1 4 10 1 0 3 4=3 ! ! D I A1. 3901 0 3 31 0 1 1 1=3 &quot; # &quot; # 210100 100 2 10 1=2 1 0 ! 23 ŒA I  D 1 2 1 0 1 0 ! 0 3=2 1 012001 0 12 001 &quot; # &quot; # 2 1 0 1 00 2 1 0 1 0 0 0 3=2 1 1=2 1 0 ! 0 3=2 0 3=4 3=2 3=4 ! 0 0 4=3 0 0 4=3 1=3 2 =3 1 1=3 2 =3 1 &quot; # &quot; # 2 0 0 3=2 1 1=2 100 3=4 1=2 1=4 0 3=2 0 3=4 3=2 3=4 ! 0 1 0 1=2 1 1=2 D 0 0 4=3 001 1=3 2 =3 1 1=4 1=2 3=4 ŒI A 1 . &quot; #&quot; #&quot; # 1ab100 1a010 b 1001 a ac b c!0100 1 c. 24 0 1 c 0 1 0 ! 0 1 0 0 1 001001 00100 1 0010 0 1    &quot; #&quot; #&quot; # &quot;# #1 211 3 1 1 2 1 1 1 0 1 1 3 1I 1 2 1 1 D 0 so B 1 does 25 1 2 1 D 4 112 1 1 3 1 1 2 1 0 not exist.          10 12 12 1 1 10 10 26 E21 A D D . E12 E21 A D AD . 21 26 02 0 1 21 02   1 0 Multiply by D D to reach DE12 E21 A D I . Then A 1 D DE12 E21 D 0 1=2   6 2 1 . 2 2 1 &quot; # &quot; # 1 0 0 2 1 0 1 1 2 1 3 (notice the pattern); A D 1 2 1. 27 A D 0 0 1 0 1 1         0210 2201 20 11 10 1=2 1=2 28 ! ! ! . 2201 0210 02 10 01 1=2 0   This is I A 1 : row exchanges are certainly allowed in Gauss-Jordan. &quot; 29 (a) True (If A has a row of zeros, then every AB has too, and AB D I is impossible) (b) False (the matrix of all ones is singular even with diagonal 1’s: ones (3) has 3 equal rows) (c) True (the inverse of A 1 is A and the inverse of A2 is .A 1 /2 /. Solutions to Exercises 20 30 This A is not invertible for c D 7 (equal columns), c D 2 (equal rows), c D 0 (zero column). 31 Elimination produces the pivots a and a b and a b . A 2 3 1 D 1 a.a b/ &quot; # a0b a a 0. 0aa 1100 60 1 1 07 32 A D4 . When the triangular A alternates 1 and 1 on its diagonal, 0 0 1 15 0001 A 1 is bidiagonal with 1’s on the diagonal and ﬁrst superdiagonal. 33 x D .1; 1; : : : ; 1/ has P x D Qx so .P Q/x D 0.       I0 A1 0 DI 34 and and . CI I0 D 1 CA 1 D 1 1 35 A can be invertible with diagonal zeros. B is singular because each row adds to zero. 36 The equation LDLD D I says that LD D pascal .4; 1/ is its own inverse. 37 hilb(6) is not the exact Hilbert matrix because fractions are rounded off. So inv(hilb(6)) 38 39 40 41 42 is not the exact either. The three Pascal matrices have P D LU D LLT and then inv.P / D inv.LT /inv.L/. Ax D b has many solutions when A D ones .4; 4/ D singular matrix and b D ones .4; 1/. Anb in MATLAB will pick the shortest solution x D .1; 1; 1; 1/=4. This is the only solution that is combination of the rows of A (later it comes from the “pseudoinverse” AC D pinv(A) which replaces A 1 when A is singular). Any vector that solves Ax D 0 could be added to this particular solution x . 3 3 2 2 1 a 0 0 1 a ab abc 1 b 07 bc 7 60 60 1 b The inverse of A D 4 is A 1 D 4 . (This 0 0 1 c5 00 1 c5 0 0 0 1 00 0 1 would be a good example for the cofactor formula A 1 D C T = det A in Section 5.3) 32 32 3 32 2 1 1 1 1 1 76 7 6a 1 7 7 60 1 6a 1 The product 4 54 5 D 4b d 1 5 5 40 d 1 1 b01 c001 0e01 f1 cef1 that in this order the multiplie...
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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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