Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# 46 inverting the identity ai c ba d i c aba gives i c

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Unformatted text preview: rs shows a; b; c; d; e; f are unchanged in the product (important for A D LU in Section 2.6). MM 1 D .In U V / .In C U.Im V U / 1 V / .this is testing formula 3/ D In U V C U.Im V U / 1 V U V U.Im V U / 1 V .keep simplifying/ D In U V C U.Im V U /.Im V U / 1 V D In .formulas 1; 2; 4 are similar/  43 4 by 4 still with T11 D 1 has pivots 1; 1; 1; 1; reversing to T  D UL makes T44 D 1. 44 Add the equations C x D b to ﬁnd 0 D b1 C b2 C b3 C b4 . Same for F x D b. CA 1 B (and d c b=a is the correct second pivot of an  ordinary 2 by 2 matrix). The example problem has     1 10 4 56 SD . 33 D 4 01 65 2 45 The block pivots are A and S D D Solutions to Exercises 21 1 A 1 D A 1 .I C AB/ 1 . So I CBA and I CAB are both invertible or both singular when A is invertible. (This remains true also when A is singular : Problem 6.6.19 will show that AB and BA have the same nonzero eigenvalues, and we are looking here at  D 1.) 46 Inverting the identity A.I C BA/ D .I C AB/A gives .I C BA/ Problem Set 2.6, page 102        10 10 x 5 1 `21 D 1 multiplied row 1; L D times D D c is Ax D b : 11 11 y 2      11 x 5 D . 12 y 7      1 0 c1 5 5 2 Lc D b is D , solved by c D as elimination goes forward. 1 1 c2 7 2       11 x 5 3 U x D c is D , solved by x D in back substitution. 01 y 2 2 3 `31 D 1 and `32 D 2 (and `33 D 1): reverse steps to get Au D b from U x D c : 4 5 6 7 8 9 1 times .x Cy Cz D 5/C2 times .y C2z D 2/C1 times .z D 2/ gives x C3y C6z D 11. " #" # " # " #" # " # "# 1 5 5 111 5 5 2D 7 ; Ux D 12 x D 2 ; xD 2. Lc D 1 1 121 2 11 1 2 2 " #" #" # 1 210 210 01 0 4 2 D 0 4 2 D U . With E 1 as L, A D LU D EA D 301 635 005 " # 1 01 U. 301 #" # " # " # " 1 111 100 1 01 21 A D 0 2 3 D U . Then A D 2 1 0 U is 021 001 006 021 the same as E211 E321 U D LU . The multipliers `21 ; `32 D 2 fall into place in L. #" #" # #" " 100 1 1 1 21 2 2 2 . This is 1 1 E32 E31 E21 A D 345 21 3 1 1 # " # " 100 101 0 2 0 D U . Put those multipliers 2; 3; 2 into L. Then A D 2 1 0 U D LU . 002 321 # #" #" " #" 1 1 1 1 a1 a 1 1 1 . D E D E32 E31 E21 D ac b c1 b 1 1 c1 The multipliers are just a; b; c and the upper triangular U is I . In this case A D L and its inverse is that matrix E D L 1 . # #" " #" 1 d e g d D 1; e D 1, then l D 1 110 f h f D 0 is not allowed 2 by 2: d D 0 not allowed; 1 1 2 D l 1 i no pivot in row 2 mn1 121 Solutions to Exercises 22 10 c D 2 leads to zero in the second pivot position: exchange rows and not singular. 11 12 13 14 15 16 c D 1 leads to zero in the third pivot position. In this case the matrix is singular. " # " # 248 2 3 A D 0 3 9 has L D I (A is already upper triangular) and D D I 007 7 " # 124 A D LU has U D A; A D LDU has U D D 1 A D 0 1 3 with 1’s on the 001 diagonal.        24 10 24 10 20 12 AD D D D LDU ; U is LT 4 11 21 03 21 03 01 " #" #" #" #" # 1 14...
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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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