Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# Key point elimination from both sides gives the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: d b2 e bc The examples 4 3 9 and b d e lead to and . 7 32 e bc f c2 890 cef " # " #" #" # " #" # 1 1 1 0 1 1 1 1 2 0 1 1; 1 AD 1 1 1 1 22 1 AD 0 1 1 231 1 1 201 1 2 3 0001 This cyclic P exchanges rows 1-2 then 61 0 0 07 23 A D 4 D P and L D U D I . 0 1 0 05 rows 2-3 then rows 3-4. 0010 " #" #" #" # 1 012 1 2 1 1 1 038D0 1 3 8 . If we wait 24 PA D LU is 1 211 0 1=3 1 2 =3 " #" #" # 1 1 211 1 0 1 2. to exchange and a12 is the pivot, A D L1 P1 U1 D 3 1 11 002 25 The splu code will not end when abs.A.k; k// < tol line 4 of the slu code on page 100. Instead splu looks for a nonzero entry below the diagonal in the current column k , and executes a row exchange. The 4 lines to exchange row k with row r are at the end of Section 2.7 (page 113). To ﬁnd that nonzero entry A.r; k/, follow abs.A.k; k// < tol by locating the ﬁrst nonzero (or the largest A.r; k/ out of r D k C 1; : : : ; n). 26 One way to decide even vs. odd is to count all pairs that P has in the wrong order. Then P is even or odd when that count is even or odd. Hard step: Show that an exchange always switches that count! Then 3 or 5 exchanges will leave that count odd. # # " " 1 100 T 31 puts 0 in the 2; 1 entry of E21 A. Then E21 AE21 D 0 2 4 27 (a) E21 D 049 1 # " 1 1 is still symmetric, with zero also in its 1, 2 entry. (b) Now use E32 D 41 T T to make the 3, 2 entry zero and E32 E21 AE21 E32 D D also has zero in its 2, 3 entry. Key point: Elimination from both sides gives the symmetric LDLT directly. 2 3 0123 61 2 3 07 28 A D 4 D AT has 0; 1; 2; 3 in every row. (I don’t know any rules for a 2 3 0 15 3012 symmetric construction like this) Solutions to Exercises 26 29 Reordering the rows and/or the columns of a b c d will move the entry a. So the result cannot be the transpose (which doesn’t move a). " #" # " # 1 0 1 yBC yBC C yBS 1 1 0 yCS yBC C yCS . 30 (a) Total currents are AT y D D 0 1 1 yBS yCS yBS (b) Either way .Ax /T y D x T .AT y / D xB yBC C xB yBS xC yBC C xC yCS xS yCS xS yBS . " #   " 700 #   1 50   x1 1 40 2 6820 1 truck T 3 31 40 1000 D Ax ; A y D D x2 50 1000 50 188000 1 plane 2 50 3000 32 Ax  y is the cost of inputs while x  AT y is the value of outputs. 33 P 3 D I so three rotations for 360ı ; P rotates around .1; 1; 1/ by 120ı . 34  1 4   2 10 1 D 9 21 2 35 L.U T /  2 D EH D (elementory matrix) times (symmetric matrix). 5 1 is lower triangular times lower triangular, so lower triangular. The transpose of U DU is U T D T U T T D U T DU again, so U T DU is symmetric. The factorization multiplies lower triangular by symmetric to get LDU which is A. T 36 These are groups: Lower triangular with diagonal 1’s, diagonal invertible D , permuta1 tions P , orthogonal matrices with QT D Q .  1  1 is southeast: 1 1 D 0 1. 10 1 The rows of B are in reverse order from a lower triangular L, so B D PL. Then B 1 D L 1 P 1 has the columns in reverse ord...
View Full Document

## 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.

Ask a homework question - tutors are online