CODEWITHSUNDEEP
matlabmatrix-inverse
e0n
e0n

Reputation: 41

Finding matrix inverse by Gaussian Elimination With Partial Pivoting

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

Upvotes: 4

Views: 6483

Answers (1)

Engineero
Engineero

Reputation: 12938

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

Upvotes: 3

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