MCS 471 Practice Problems for Computational Linear Algebra with Some Answers ------------------------------------------------------------------------------- CAUTION: These problems are just for practice and are not to be handed in. Some of these old exam problems and answers may not be relevant to your current course. Caveat Usor! In the following problems, use "EXAM PRECISION: chop to given number of significant digits only when you write an intermediate or final answer down and continue calculations with those numbers recorded. ------------------------------------------------------------------------------- 1. Solve: 0.543E-3*X(1) + 3.21*X(2) = 3.87 4.32 *X(1) + 2.31*X(2) = 4.92 using only forward Gaussian elimination with back substitution to 3 significant digits. Record multipliers. Use no partial pivoting nor use scaling. (Final ans.: X =(33.1,1.20)T). 2. Solve: 0.543*X(1) + 3.21e3*X(2) = 3.87e3 4.32 *X(1) + 2.31 *X(2) = 4.92 using only forward Gaussian elimination With "virtual" partial pivoting and back substitution to 3 significant digits. Record multipliers and pivot vectors for each elimination step; but use no scaling. (Final ans.: X=(0.497, 1.20)T). 3. Solve: 0.995*X(1) + 1.54 *X(2) + 4.51*X(3) = 43.1 0.995*X(1) + 2.16 *X(2) + 1.19*X(3) = 31.6 0.298*X(1) + 0.577*X(2) + 1.42*X(3) = 16.2 using forward Gaussian elimination with "virtual" partial pivoting, "virtual" scaling and back substitution chopping to 3 significant digits. Show multipliers, scale vectors and pivot vectors for each step. Record the final answer and that only rounded to 2 significant digits. (Final ans.: X =(4c) (-29.1,23.7,7.89) =(2r) (-29.,+24.,+7.9)T ). 4. Solve: 8.955*X(1) + 9.230*X(2) + 13.53*X(3) = 63.4 4.479*X(1) + 5.770*X(2) + 7.058*X(3) = 79.6 8.955*X(1) + 12.31*X(2) + 3.530*X(3) = 95.5 using forward Gaussian elimination, "virtual" scaling, "virtual" partial pivoting and back substitution to 4 significant digits. Record augmented matrices with scale and pivot vectors at each elimination step. Round your final solution to 3 significant digits. (Ans.: 8.955 9.230 13.53 63.40 1 13.53 (A/B/P/S)= 4.479 5.770 7.058 79.60 2 7.058 8.955 12.31 3.530 95.50 3 12.31 (+1.000)m -3.080 10.00 -32.10 3 10.00 =(4c) (+.5001)m -.3862 5.292 31.84 2 5.292 8.955 12.31 3.530 95.50 1 ----- ----- -3.080 10.00 -32.10 3 ----- =(4c) ----- (+.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 5. Solve A*X=B and simultaneously find the inverse A**(-1) for 6.245e-5*X(1)+3.872*X(2)=3.741 2.236 *X(1)+5.292*X(2)=5.099 using forward Gaussian elimination with back substitution only. (Note: use no pivoting or scaling.) Compute the condition number of A in the 1-norm. (partial ans.: X=(4c) 10.37; A**(-1)=(4c) 3.996 .4473; .9960 .2582 -7.215e-6 ||A**(-1)||1 =(4c) 4.254; Cond(A)1=(4c) 38.98 6. Using Newton's method and forward Gaussian elimination, approximate the vector zero of f1(X,Y) = 4*X**2+Y**2-4 f2(X,Y) = exp(Y)-X-2 , by starting at (X,Y)(1) = (1., 1.) and finding the next iterate as well as its vector-f and its gradient. (partial ans.: (X,Y)=(4c) (.8619,1.052); (f1,f2)=(4c) (.07819,.001472); grad(f)=(6.895,2.104,-1.,2.863) by cols. (alternate ans.: (X,Y)=(4c) (.8618,1.052); (f1,f2)=(4c)(-.07750,.001572); grad(f)=(6.894,2.104,-1.,2.863) by columns. 7(me/a/w83). Approximate the solution of the following linear algebraic system using forward Gaussian elimination, partial pivoting with "virtual" scaling (i.e., not actual scaling) and back substitution. At each elimination step record the augmented matrix (A/B), pivot vector, scale vector and multipliers chopped to 4 significant digits and continue calculations with these recorded results. Calculate the absolute value norm of the residuals relative to the approximate solution in the same norm. 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) .4043e-3 . 8(me/a/s82). Approximate the solution of the following linear algebraic system using forward Gaussian elimination along with "virtual" scaling, partial pivoting and back substitution. At each elimination step record the augmented matrix (A/B), scale vector, pivot vector and multipliers chopped to 4 significant digits and continue calculations with these chopped results and the number of multiplications or divisions and the number of additions or subtractions used. 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 (Ans.: 4.477 1.538 13.53 31.72 1 13.53 (A/B/P/S) = 8.958 3.846 28.23 159.2 2 28.23 8.955 4.103 7.063 95.49 3 8.955 .4999m -.5130 9.999 -16.01 3 9.999 =(4c) 1.000m -.2570 21.16 63.71 2 21.16 8.955 4.103 7.063 95.49 1 ----- ----- -.5130 9.999 -16.01 3 ----- ; =(4c) ----- .5009m 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 -------------------------------------------------------------------------------