Problem #1. Set n 200 and generate an n × n matrix and two vectors in Rn (all ha

Question

Problem #1. Set n 200 and generate an n × n matrix and two vectors in Rn (all having integer entries) via the following commands A = floor (10*rand(n)); b = sum(A,2); z = ones (n ,1); Since this matrix and these vectors are large, you want semicolons here to suppress the output. (a) The exact solution to Ax -b should be the vector z. Explain why. You can solve the linear system in MATLAB via either the backslash operation or by forming A-1b. What is the difference? Write a script to both time and compute the errors for each approach.1 Include the commands above and use tic toc as follows. tic; x = A; t1 = toc ; err1 max (abs (x-z)); disp(['time for backslash solve: ',num2str(t1)]) disp(['error for backslash solve: ',num2str(erri)]) tic; y inv (A) *b; t2 toc ; err2 = max (abs (y-z)); disp(['time for inverse A solve: ',num2str (t2)1) disp(['error for inverse A solve: ',num2str (err2)]) Which is faster and which has better error? Can you explain the results? (b) Repeat using n = 500 and n = 1000. Include your script, the output of your script, and comments on what you observe.

Explanation / Answer

Solution:

Consider the data set n = 200 and generate a n x n matrix and two vectors in Rn

both having integer entries using MATLAB as follows:
>> n=200;

>> A=floor(10*rand(n));

>> b=sum(A')';

>> z=ones(n.1);

a)

(a) The exact solution of the system Ax = b should be the vector z because in the rows of the matrix b.

we add the elements of the corresponding rows of matrix A and so the solution is 11 x I matrix with all entries as 1.

Now, measure the elapsed time by using the following MATLAB commands:

Input: >> tic,x=A1b;toc

Output: Elapsed time is 0.783085 seconds.

Input: >> tic,y=inv(A)*b,toc

Output: Elapsed time is 0.084168 seconds.

From the time calculation, we observe that the method using "I" operation for the system

Ax = b is faster than the method of computing the inverse.

Now, compare the accuracy of the two methods by using the following code in Matlab:

Input: >> max(abs(x-z))

Output: ans = 3.3273e-13

Input: >>max(abs(y-z))

Output: ans = 2.0428e-12

From the above calculation, we can see that the accurate solutions are same.

(b)

For n = 500
>> n=500;

>> A=floor(10trand(n));

>> b=sum(A')';

>> z=ones(n.1);

We can measure the elapsed time by using the following Matlab command:

Input: >>tic,x=A1b;toc

Output: Elapsed time is 0.033196 seconds.

Input: >> tic,y=inv(A)*b;toc

Output: Elapsed time is 0.061206 seconds.

From the elapsed time calculations, observed that the method using "1" operation for the system Ax = b is faster than the method of computing the inverse.

Now. we compare the accuracy of the two methods by using the following code in Matlab:

Input: >> max(abs(x-z))

Output: ans =

6.2839e-13

Input: >> max(abs(y-z))

Output: ans =

3.4674e-12

From the above calculation. observed that the method using "" operation give more accurate solution than the method of computing the inverse

For n =1000

>> n=1000;

>> A=floor(10*rand(n));

>> b=sum(A')';

>> z=ones(n,1);

We can measure the elapsed time by using the following Matlab command:

Input: >> tic,x=A;toc

Output: Elapsed time is 0.143390 seconds.

Input: >> tic,y=inv(A)*b;toc

Output: Elapsed time is 0.293533 seconds.

From the elapsed time calculation, observed that the method using "' operation for the system Ax = b is faster than the method of computing the inverse.

Now. compare the accuracy of the two methods by using the following code in Matlab:

Input:

>> max(abs(x-z))

Output:

ans =

5.2336e-12

Input:

>> max(abs(y-z))

Output: ans =

3.5584e-11
From the above calculation. observe that the method using "" operation give more accurate solution than the method of computing the inverse

I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)

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