Finding matrix inverse by Gaussian Elimination With Partial Pivoting - matlab

Question

Hello guys I am writing program to compute determinant(this part i already did) and Inverse matrix with GEPP. Here problem arises since i have completely no idea how to inverse Matrix using GEPP, i know how to inverse using Gauss Elimination ([A|I]=>[I|B]). I have searched through internet but still no clue, could you please explain me?
Here is my matlab code (maybe someone will find it useful), as of now it solves AX=b and computes determinant:
function [det1,X ] = gauss_czesciowy( A, b )
%GEPP
perm=0;
n = length(b);
if n~=m
error('vector has wrong size');
end
for j = 1:n
p=j;
% choice of main element
for i = j:n
if abs(A(i,j)) >= abs(A(p,j))
p = i;
end
end
if A(p,j) == 0
error('Matrix A is singular');
end
%rows permutation
t = A(p,:);
A(p,:) = A(j,:);
A(j,:) = t;
t = b(p);
b(p) = b(j);
b(j) = t;
if~(p==i)
perm=perm+1;
end
% reduction
for i = j+1:n
t = (A(i,j)/A(j,j));
A(i,:) = A(i,:)-A(j,:)*t;
b(i) = b(i)-b(j)*t;
end
end
%determinant
mn=1;
for i=1:n
mn=mn*A(i,i);
end
det1=mn*(-1)^perm;
% solution
X = zeros(1,n);
X(n) = b(n)/A(n,n);
if (det1~=0)
for i = 1:n
s = sum( A(i, (i+1):n) .* X((i+1):n) );
X(i) = (b(i) - s) / A(i,i);
end
end
end

Answer 1

Here is the algorithm for Guassian elimination with partial pivoting. Basically you do Gaussian elimination as usual, but at each step you exchange rows to pick the largest-valued pivot available.
To get the inverse, you have to keep track of how you are switching rows and create a permutation matrix P. The permutation matrix is just the identity matrix of the same size as your A-matrix, but with the same row switches performed. Then you have:
[A] --> GEPP --> [B] and [P]
[A]^(-1) = [B]*[P]
I would try this on a couple of matrices just to be sure.
EDIT: Rather than empirically testing this, let's reason it out. Basically what you are doing when you switch rows in A is you are multiplying it by your permutation matrix P. You could just do this before you started GE and end up with the same result, which would be:
[P*A|I] --> GE --> [I|B] or
(P*A)^(-1) = B
Due to the properties of the inverse operation, this can be rewritten:
A^(-1) * P^(-1) = B
And you can multiply both sides by P on the right to get:
A^(-1) * P^(-1)*P = B*P
A^(-1) * I = B*P
A^(-1) = B*P