Create a function in MATLAB that finds the unique solution to Ax=b equation usin

Question

Create a function in MATLAB that finds the unique solution to Ax=b equation using Gaussian elimination with partial pivoting where A is a square matrix and b is a column vector of the same height. The function will take A and b as inputs and give x as the output. Partial pivoting is choosing the largest absolute value possible for a diagonal entry within the active column and changing rows accordingly.

Part 1: Start by writing a function (with comments) to achieve these: Determine if the matrix A is square. Augment A and b. Find the maximum absolute value in the first column and the row number that this belongs to using a forloop. Interchange the first row with the row with the maximum firstcolumnentry. (To achieve this, create a short subfunction that interchanges two specified rows for any matrix.) Submit a single mfile with a subfunction that switches rows

Explanation / Answer

Steps:

1. Find the entry in the left column with the largest absolute value. This

entry is called the pivot.

2. Interchange rows, if required , so that the pivot is in the 1st row.

3. Divide the 1st row by the pivot. (This step is unnecessary if the pivot is 1)

4. Use elementary row operations to reduce the remaining entries in the

first column to zero.

The completion of these four steps is called a pass. After performing the

first pass, ignore the first row and first column and repeat the four steps

on the remaining submatrix. Continue this process until the matrix is in

row-echelon form.

Code:

function gaussian_partial

  A=input('provide matrix A:');

   b=input('provide matrix b:');

   gauss(A,b)

end

function x=gauss(A,b)

n = size(A,1); %getting n

A = [A,b];      %produces the augmented matrix

%elimination process starts

for i = 1:n-1

   p = i;

   %comparison to select the pivot

   for j = i+1:n

       if abs(A(j,i)) > abs(A(i,i))

           U = A(i,:);

           A(i,:) = A(j,:);

           A(j,:) = U;       

       end

   end

   %checking for nullity of the pivots

   while A(p,i)== 0 & p <= n

       p = p+1;

   end

   if p == n+1

       disp('No unique solution');

       break

   else

       if p ~= i

           T = A(i,:);

           A(i,:) = A(p,:);

           A(p,:) = T;

       end

   end

   for j = i+1:n

       m = A(j,i)/A(i,i);

       for k = i+1:n+1

           A(j,k) = A(j,k) - m*A(i,k);

       end

   end

end

%checking for nonzero of last entry

if A(n,n) == 0

   disp('No unique solution');

   return

end

%backward substitution

x(n) = A(n,n+1)/A(n,n);

for i = n - 1:-1:1

   sumax = 0;

   for j = i+1:n

       sumax = sumax + A(i,j)*x(j);

   end

   x(i) = (A(i,n+1) - sumax)/A(i,i);

end

end

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