Chapter 7, Problem 1E

Set
$\begin{array}{l}A=round\left(10*rand\left(6\right)\right)\\ s=ones\left(6,1\right)\\ b=A*s\end{array}$
The solution of the linear system $Ax=b$ is clearly s. Solve the system using the MATLAB \ operation. Compute the error $x-s.$ (Since s consists entirely of I's, this is the same as $x-1.$ ) Now perturb the system slightly. Set
$\begin{array}{l}t=1.0e-12,\\ E=rand\left(6\right)-0.5\\ r=rand\left(6,1\right)-0.5\end{array}$
and set
$M=A+t*E,c=b+t*r$
Solve the perturbed system $Mz=c$ for z. Compare the solution z to the solution of the original system by computing $z-1.$ How does the size of the perturbation in the solution compare with the size of the perturbations in A and b? Repeat the perturbation analysis with $t=1.0e-04$ and $t=1.0e-02.$ Is the system $Ax=b$ well conditioned? Explain. Use MATLAB to compute the condition number of A.

To determine

Calculate the eigenvector, error and the condition number of A.

Answer to Problem 1E

The solution is

  $x=\left[\begin{array}{l}\text{1}\\ \text{1}\\ \text{1}\\ \text{1}\\ \text{1}\\ \text{1}\end{array}\right]$

  $error=x-s=\left[\begin{array}{l}\text{ 0}\\ \text{-0}\text{.5551}\\ \text{ 0}\text{.6661}\\ \text{-0}\text{.3331}\\ \text{ 0}\text{.4441}\\ \text{-0}\text{.3331}\end{array}\right]\times {10}^{-15}$

Solution of the perturbed system is

  $z=\left[\begin{array}{l}\text{1}\\ \text{1}\\ \text{1}\\ \text{1}\\ \text{1}\\ \text{1}\end{array}\right]$

  $z-s=\left[\begin{array}{l}\text{ 0}\text{.1175}\\ \text{ 0}\text{.0462}\\ \text{ 0}\text{.7292}\\ \text{-0}\text{.1156}\\ \text{-0}\text{.4740}\\ \text{-0}\text{.6285}\end{array}\right]\times {10}^{-12}$

The condition number of A is

  $ConditionnoofA=\text{46}\text{.015}$

Explanation of Solution

Given: The matrix has been given

  $\begin{array}{l}A=round(10\times rand(6))\\ s=ones(6,1)\\ b=A\times S\\ t={10}^{-12}\\ E=rand(6)-0.5\\ r=rand(6,1)-0.5\\ M=A+t\times E\\ C=b+t\times r\end{array}$

Concept Used:

Given,

  $\begin{array}{l}A=round(10\times rand(6))\\ s=ones(6,1)\\ b=A\times S\end{array}$

To calculate the eigenvector, we use

  $\begin{array}{l}Ax=b\\ x=b/A\end{array}$

Then compute the error from s vector

  $error=x-s$

Calculate the perturb solution using given relation

  $\begin{array}{l}E=rand(6)-0.5\\ r=rand(6,1)-0.5\\ M=A+t\times E\\ C=b+t\times r\end{array}$

Compute the error from s vector

  $z-s=\left[\begin{array}{l}\text{ 0}\text{.1175}\\ \text{ 0}\text{.0462}\\ \text{ 0}\text{.7292}\\ \text{-0}\text{.1156}\\ \text{-0}\text{.4740}\\ \text{-0}\text{.6285}\end{array}\right]\times {10}^{-12}$

Calculate the condition number using function “cond(A)” in MATLAB

  $ConditionnoofA=\text{46}\text{.015}$

Program:

clc
clear
close all
A = round(10 * rand(6));
s = ones(6, 1);
b = A * s;
x = A \ b;
fprintf('Solution to the system:\n');
disp(x);
err = x - s;
fprintf('Error:\n');
disp(err);
t = 1.0e-12;
E = rand(6) - 0.5;
r = rand(6, 1) - 0.5;
M = A + t * E;
c = b + t * r;
z = M \ c;
fprintf('Solution to the perturbed system for t = %.12f:\n', t);
disp(z);
err = z - s;
fprintf('Difference between the solutions:\n');
disp(err);
t = 1.0e-04;
E = rand(6) - 0.5;
r = rand(6, 1) - 0.5;
M = A + t * E;
c = b + t * r;
z = M \ c;
fprintf('Solution to the perturbed system for t = %f:\n', t);
disp(z);
err = z - s;
fprintf('Difference between the solutions:\n');
disp(err);
t = 1.0e-02;
E = rand(6) - 0.5;
r = rand(6, 1) - 0.5;
M = A + t * E;
c = b + t * r;
z = M \ c;
fprintf('Solution to the perturbed system for t = %f:\n', t);
disp(z);
err = z - s;
fprintf('Difference between the solutions:\n');
disp(err);
fprintf('Condition number of A = %f', cond(A));

Quarry:

• First, we have defined the given matrix.
• Then compute the eigenvector.
• Calculate the error.
• Calculate the perturb solution at the given time interval.
• Calculate the condition vector of A.