Question

Engineering Computations and Modelling > Week 6 Homework > Backward Substitution 2 solutions submitted (max: Unlimited) | View my solutions Problem Summary Consider the matrix problem Ux=b, where U is an upper triangular matrix and b is a column vector with the form: U11 U12 U13 U in bi 0 Uan b2 U22 U23 0 U33 U= 0 Uzn b= |b3 . ... ... ... ... 0 0 0 bn This problem can be solved using a simple algorithm called backward substitution. In this algorithm, xn is easily solved from the last equation in the matrix problem. This result is then substituted into the second to last equation, which is then explicitly solved for xn-1. This process is repeated until all values of x are found. Complete the definition of the backsub function below by implementing the backward substitution algorithm. This function takes as input: U, an nxn upper triangular matrix b, an nx 1 vector for the right-hand-side of the matrix equation

HERE IS THE CODE I FIXED BUT STILL DOESN'T WORK

NOTE: THE VARIABLE x = zeros(size(b)) CAN'T BE CHANGED CAUSE HAS BEEN SET BY ASSESSOR

HI EXPERTS I NEED HELP TO SOLVE THIS HOMEWORK PROBLEM FOR MATLAB CODE

A COUPLE OF TIMES I TRIED LAST TIME TO ASK BUT ALL OF THE ANSWERS WERE WRONG

PLEASE KINDLY HELP ME FIND THE RIGHT SOLUTION, ANY HELP WILL BE VERY APPRECIATE

Answer 1

Ok, your algorithm is a bit off the mark. You were thinking along the right paths, however. For these kinds of problems, the best way to learn is to do the steps by hand and then thinking about how you will implement them via code.

I have provided both the text and the screenshot for your better understanding. The screenshot also has some comments for your help. Please note that this is a script file which is to be saved as backsub.m in your MATLAB root directory.
The script backsub.m gives the value of the variable matrix as output. The input must be a square upper triangular non-singular matrix with none of the diagonal elements as 0.

backsub.m

+ backsub.m function x = backsub (U,b) n = length(b); Scalculating size of solution vector x = ones (size (b)); $initializing solution vector 7 - for i=n:-1:1 x(i) = b(i)/U(i,i); $solving the last row of the matrix b(1:1-1) = 5(1:1-1) - U(1:1-1,i)*x(i); updating the constant values so that the previous row can now be treated as the last row of the matrix end fprintf("result of backward substitution\n"); disp (x); end 10 11 - TIT 12 -

function x = backsub(U,b)

n = length(b); %calculating size of solution vector
x = ones(size(b)); %initializing solution vector

for i=n:-1:1
x(i) = b(i)/U(i,i); %solving the last row of the matrix
b(1:i-1) = b(1:i-1) - U(1:i-1,i)*x(i); %updating the constant values so that the previous row can now be treated as the last row of the matrix
end
fprintf("result of backward substitution\n");
disp(x);
end

OUTPUT

I have attached screenshots for 3 random test cases, one of them in the question and the other 2 from random sources.

Example matrix in question

>> A = [1,2;0,2] 2 1 2 0 >> B = [2;1] B = 1 >> x = backsub (A,B); result of backward substitution 1.0000 0.5000 fx >> |

3x3 matrix

>> A = [1,-2,1;0,1,6;0,0,1] 1 6 -2 1 0 >> b = [ 4;-1;2] b = 4 2 >> x = backsub (A,b) result of backward substitution -24 -13 2

4x4 matrix

>> A = [ 1,2,1,-1 ; 0,-4,1,7;0,0,-2,1;0,0,0,-1] 1 1 -1 7 1 1 2 0 -4 0 -2 0 0 0 >> B = [5;1;1;3] B = 5 -1 1 1 3 >> x = backsub (A,B); result of backward substitution 16 -6 -2 -3

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