6.684*x_1 + 2.925*x_2 + 9.835*x_3 = 10.75
5.543*x_1 + 5.953*x_2 + 88.63*x_3 = 19.81
8.375*x_1 + 2.988*x_2 + 8.681*x_3 = 24.72
Record augmented matrices with marked multipliers, pivot vectors, and scale vectors, i.e., [A | b || P || S ], at each elimination step, including the initial step.
[ A | b || P || S ] (a_ik/S_i) | 6.684 2.925 9.835 | 10.75 || 1 || 9.835 | (0.6796) =(4R) | 5.543 5.953 88.63 | 19.81 || 2 || 88.63 | (0.06254) | 8.375 2.988 8.681 | 24.72 || 3 || 8.681 | (0.9648)max |(0.7981)m 0.5403 2.907 | -8.979 || 3 || 2.907 | (0.1859)max ~(4R) |(0.6619)m 3.975 82.88 | 3.448 || 2 || 82.88 | (0.04796) fge1 | 8.375 2.988 8.681 | 24.72 || 1 || 8.681 | (------) |(0.7981)m 0.5403 2.907 | -8.979 || 3 || 2.907 | (------) ~(4R) |(0.6619)m (7.357)m 61.49 | 69.51 || 1 || 82.88 | (------) fge2 | 8.375 2.988 8.681 | 24.72 || 2 || 8.681 | (------) Fast BackSubstitution in Single Calculations => | 9.879 | X =(4R) | -22.70 | | 1.130 | Slow BackSubstitution with Imediate Substitution of Each Component => | 9.876 | X =(4R) | -22.69 | | 1.130 |
| 1 0 0 | | 8.375 2.988 8.681 | L_ap =(4R) | 0.7981 1 0 |, U_ap =(4R) | 0 0.5403 2.907 | | 0.6619 7.357 1 | | 0 0 61.49 |
w^{(0)} = [ -0.3 0.9 -0.04 ]^{T}.
Use the infinity norm. {Caution: Do NOT compute any inverses directly, else there will be a large deduction of points for inefficiency. Ignore the fact that Aap is not a symmetric matrix.}
||w^{(0)}||_{inf} = 0.9; \lambda_3^{(0)}=1/||w^{(0)}||_{inf} =(4R) +1.111; [ -0.3333 ] X_3^{(0)}=what^{(0)}=w^{(0)}/||w^{(0)}||_{inf} =(4R) [ 1.000 ]; [ -0.04444 ] [ -0.3333 ] L_ap*y^{(1)}=what^{(0)}; y^{(1)} =(4R) [ 1.266 ] = (4R) U_ap·a;w^{(1)}; [ -9.138 ] [ -1.007 ] w^{(1)} =(4R) [ 3.143 ]; \lambda_3^{(1)}=1/||w^{(1)}||_{inf} =(4R) +0.3182; [ -0.1486 ] [ -0.3204 ] X_3^{(1)}=what^{(1)}=w^{(1)}/||w^{(1)}||_{inf} =(4R) [ 1.000 ]; [ -0.04728 ]
|| e ||_{p}/|| x ||_{p} < Cond_{p}[A]·|| r ||_{p} / || b ||_{p}.
Give reasons for every step. {You may assume that A·e=r is known, that A^{-1} exists, and || x ||_{p} not= 0 & || b ||_{p} not= 0.}
Since A^{-} exists and A*e=r, e = A^{-1}*r. By Cauchy's inequality, ||e||_{p} < ||A^{-1}}||_{p}*||r||_{p} (#1). Since A*x = b by definition and Cauchy's equality again, ||b||_{p} < ||A||_{p}*||x||_{p}, so 1/||x||_{p} < ||A||_{p} / ||b||_{p} (#2). By definition, cond_{p}[A] = ||A||_{p}*||A^{-1}||_{p}, so by using (#1) & (#2), ||e||_{p}/||x||_{p} < ||A^{-1}||_{p}*||r||_{p}/(||b||_{p} / ||A||_{p}) = cond_{p}[A]*||r||_{p}/||b||_{p} . Note that by assumption, both divisors ||x||_{p} and ||b||_{p} are nonzero, and since ||b||_{p} < ||A||_{p}*||x||_{p}, ||A||_{p} can not be zero if ||b||_{p} be zero, so the additional division by ||A||_{p} in (#2) is legal. Q.E.D.
In the following older problems prior to Fall 1999, use "CHOPPING EXAM PRECISION": The answers are calculated for chopping to 4 significant (4C) digits since the problems are from a time when chopping was used.
Note: Maple comments are not part of these sample exam problems, but were added afterward in the editing stage to aid in analyzing the problems.
(Ans.: [A|B|P|S]= 8.955 9.230 13.53 63.40 1 13.53 4.479 5.770 7.058 79.60 2 7.058 8.955 12.31 3.530 95.50 3 12.31 =(4c) (+1.000)m -3.080 10.00 -32.10 3 10.00 (+.5001)m -.3862 5.292 31.84 2 5.292 8.955 12.31 3.530 95.50 1 ----- =(4c) ----- -3.080 10.00 -32.10 3 ----- ----- (+.1253)m 4.039 35.86 1 ----- , 8.955 12.31 3.530 95.50 2 ----- X(3) =(4c) 35.86/4.039 =(4c) 8.878 =(3r) 8.88 X(2) =(4c) (-31.2-10.*8.878)/(-3.08) =(4c) 39.24 =(3r) 39.2 X(1) =(4c) (95.5-3.53*8.878-12.31*39.24)/8.955 =(4c) -46.77 =(3r) -46.8
(partial (Corrected in red) ans.: X=(4c) [10.37 .9960]^{T}; A^{-1}=(4c) 3.996 0.4473e-1; 0.2582 -7.215e-7 ||A^{-1}||_{1}=(4c) 4.254; Cond(A)_{1}=(4c) 38.98
(partial ans.: (X,Y)=(4c) (.8619,1.052); (f1,f2)=(4c) (.07819,.001472); jacobian(f)=(6.895,2.104,-1.,2.863) by cols.; (alternate ans.: (X,Y)=(4c) (.8618,1.052); (f1,f2)=(4c)(-.07750,.001572); jacobian(f)=(6.894,2.104,-1.,2.863) by columns.
8.955*X(1) + 9.230*X(2) + 13.53*X(3) = 63.40 4.479*X(1) + 5.770*X(2) + 7.058*X(3) = 79.60 8.955*X(1) + 12.31*X(2) + 3.530*X(3) = 92.87 (partial ans.: X =(4c) (-47.08,39.18, 9.121); norm of rel. residuals = ||r||/||a||/||b|| =(4c) .9997e-3 , or = ||r||/||b|| =(4c) 0.4043e-3 .
4.477*X(1) + 1.538*X(2) + 13.53*X(3) = 31.72 8.958*X(1) + 3.846*X(2) + 28.23*X(3) = 159.2 8.955*X(1) + 4.103*X(2) + 7.063*X(3) = 95.49 (Answer: [A|B|P|S] = 4.477 1.538 13.53 31.72 1 13.53 8.958 3.846 28.23 159.2 2 28.23 8.955 4.103 7.063 95.49 3 8.955 =(4c) (0.4999)m -0.5130 9.999 -16.01 3 9.999 (1.000)m -0.2570 21.16 63.71 2 21.16 8.955 4.103 7.063 95.49 1 ----- =(4c) ----- -0.5130 9.999 -16.01 3 ----- ; ----- (0.5009)m 16.15 71.72 1 ----- ; 8.955 4.103 7.063 95.49 2 ----- ; X(3) =(4c) 4.440 =(3r) 4.44 X(2) =(4c) 117.7 =(3r) 118. X(1) =(4c) -46.76 =(3r) -46.8 sum-mults. = 19 (or 22, depending on what is counted); sum-adds = 11
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