MCS 471 Practice Problems 2: Computational Linear Algebra

Background Reading:

• Gerald and Wheatley, Chapter 2
• Class Octave and Maple Web Pages:
  • Octave Class Examples for Solving Linear Algebra Equations.
  • Maple Class Examples for Solving Linear Algebra Equations.
  • MSCS Introduction to Maple
  • MCS 471 Beginner's Introduction to Maple in MathLab
  • Victoria Bush, Linear Algebra Using Maple (Click Here for HTML Page), Center for Statistical and Mathematical Computing, Indiana University at Bloomington, 1995.
  • MSCS Maple Plot Lab
  • MCS 471 Help for Maple Symbolic Tools

  Topics

  1. Forward Gaussian Elimination with Back Substitution
  2. Virtual Row (Partial) Pivoting and Row Scaling
  3. Virtual Full Pivoting
  4. Linear Algebra Computational Complexity
  5. Determinants, Inverse and Multiple RHSs by FGE
  6. LU Decomposition with and without Pivoting
  7. Norms, Condition Numbers and Error Propagation
  8. Multidimensional Newton's Method (Nonlinear Algebra)

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  Practice Problems:

  In computational problems, use "EXAM PRECISION": Chop to 4 significant (4C) digits only when you write an intermediate or final answer down and continue calculations with those numbers recorded.

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  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.:  [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
     
     

  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    .9960]^T;                 
     
     A^(-1)=(4c) 
     3.996   0.4473;
     0.2582  -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 Jacobian (matrix of gradients).

     (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.
     
     

  7. (masters-exam/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)  0.4043e-3 .
     
     

  8. (masters-exam/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
     
     (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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