Question

4. For n = 50, 100, 200, 300, 400, do the following. Let A be the n × n matrix with entries Aij = p 2(i − j) 2 + n/5. Define xexact = [1, 1, . . . , 1]T , and then compute b = Axexact. Using Matlab’s backslash, numerically solve the equation

Ax = b,

thereby producing a computed solution xc. In exact arithmetic xc = xexact of course, but in IEEE double precision xc will not equal xexact. The quantity kxc − xexactk∞ is the forward error. Construct a table which, for each n, collects the relative forward error, relative backward error, magnification factor, and (infinity-norm) condition number of A. Discuss your results.

Answer 1

ANSWER :

format long
n=input('Enter the size of matrix: ')
A=zeros(n,n)
for i=l:n
    for j=l:n
        A(i,j)=(2*(i-j)^2+n/5)^(1/2)
    end
end
x_exact=ones(n,l)
b=A*x_exact
x=A/b
bl=A*x
REF=norm((x-x_exact),inf)/norm(x_exact,inf)    %Relative forward Error
RBE=norm((b-bl),inf)/norm(b,inf)               %Relative Backward Error
MF=REF/RBE                                     %Magnification Factor
k=cond(A,inf)                                  %Condition number with infinity norm

N	Relative forward Error	Relative Backward Error	Magnification Factor	Condition number with infinity norm
50	6.0513*10^-11	6.499*10^-16	9311*10^4	9.82*10^5
100	3.549*10^-9	6.463*10^-16	549*10^6	6.184*10^7
200	9.3077*10^-7	5.157*10^-16	1.804*10^9	1.29*10^10
300	4.639*10^-5	1.259*10^-15	3.683*10^10	6.2039*10^11
400	8.5488*10^-4	1.2876*10^-15	6063*10^11	1.475*10^13

• Relative Farword Error is increasing as n increases.
• Relative backward Error lies between 10^-16 and 10^-15 range as n increases.
• The magnification Factor increases with increase in the value of n.
• The condition number also increases with n which means the matrix is getting more all conditioned as n increases.

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