Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# A complete graph has 5 c 4 c 3 c 2 c 1 d 15 edges

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Unformatted text preview: 3 by 3 1 “element matrix” c1 E1 D Œ 1 0 0 T c1 Œ 1 0 0  has c1 in the top left corner. 6 Multiply AT C1 A1 1 7 For 5 springs and 4 masses, the 5 by 4 A has two nonzero diagonals: all ai i D 1 and ai C1;i D 1. With C D diag.c1 ; c2 ; c3 ; c4 ; c5 / we get K D AT CA, symmetric tridiagonal with diagonal entries Ki i D ci C ci C1 and off-diagonals Ki C1;i D ci C1 . With C D I this K is the 1; 2; 1 matrix and K.2; 3; 3; 2/ D .1; 1; 1; 1/ solves K u D ones.4; 1/. (K 1 will solve K u D ones.4/.) 3 1 u00 D 1 with u.0/ D u.1/ D 0 is u.x/ D 2 .x x 2 /. At x D 1 ; 2 ; 5 ; 4 55 5 2 this gives u D 2; 3; 3; 2 (discrete solution in Problem 7) times .x/ D 1=25. 8 The solution to 9 u 00 D mg has complete solution u.x/ D A C Bx 1 mgx 2 . From u.0/ D 0 we 2 get A D 0. From u 0 .1/ D 0 we get B D mg . Then u.x/ D 1 mg.2x x 2 / at 2 12 x D 3 ; 3 ; 3 equals mg=6; 4mg=9; mg=2. This u.x/ is not proportional to the discrete 3 u D .3mg; 5mg; 6mg/ at the meshpoints. This imperfection is because the discrete problem uses a 1-sided difference, less accurate at the free end. Perfect accuracy is recovered by a centered difference (discussed on page 21 of my CSE textbook). 10 (added in later printing, changing 10-11 below into 11-12). The solution in this ﬁxed- ﬁxed case is .2:25; 2:50; 1:75/ so the second mass moves furthest. 11 The two graphs of 100 points are “discrete parabolas” starting at .0; 0/: symmetric around 50 in the ﬁxed-ﬁxed case, ending with slope zero in the ﬁxed-free case. 12 Forward/backward/centered for du=dx has a big effect because that term has the large coefﬁcient. MATLAB: E D diag(ones.6; 1/; 1/; K D 64  .2  eye.7/ E E 0 /; D D 80  .E eye.7//; .K C D/nones.7; 1/; % forward; .K D 0 /nones.7; 1/; % backward; .K C D=2 D 0 =2/nones.7; 1/I % centered is usually the best: more accurate Problem Set 8.2, page 428 1 AD &quot; 1 1 0 1 0 1 # &quot;#&quot;# 0 c 1 1 ; nullspace contains c ; 0 is not orthogonal to that nullspace. 1 c 0 2 AT y D 0 for y D .1; 1; 1/; current along edge 1, edge 3, back on edge 2 (full loop). Solutions to Exercises 3 Elimination on b1 Œ A &quot; 83 b &quot; D 1 1 0 1 0 b1 0 1 b2 1 1 b3 # leads to Œ U c D # 1 1 0 b1 0 1 1 b2 b1 . The nonzero rows of U come from edges 1 and 3 0 0 0 b3 b2 C b1 in a tree. The zero row comes from the loop (all 3 edges). 4 For the matrix in Problem 3, Ax D b is solvable for b D .1; 1; 0/ and not solvable for b D .1; 0; 0/. For solvable b (in the column space), b must be orthogonal to y D .1; 1; 1/; that combination of rows is the zero row, and b1 b2 C b3 D 0 is the third equation after elimination. 5 Kirchhoff’s Current Law AT y D f is solvable for f D .1; 1; 0/ and not solvable for f D .1; 0; 0/; f must be orthogonal to .1; 1; 1/ in the nullspace: f1 C f2 C f3 D 0. &quot; # &quot;# &quot; # &quot;# 2 1 1 3 1 c T 1 2 1 xD 3 D f produces x D 1 C c ; potentials 6 A Ax D 1 1 2 0 0 c x D 1; 1; 0 and currents Ax D 2, 1, 1; f sends 3 units from node 2 into node 1. # &quot; # &quot;# &quot; # &quot;# &quot;...
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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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