Performing Gauss Elimination with MatLab

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Lukumon Kazeem
Lukumon Kazeem am 11 Jul. 2012
Kommentiert: Walter Roberson am 25 Nov. 2024
K =
-0.2106 0.4656 -0.4531 0.7106
-0.6018 0.2421 -0.8383 1.3634
0.0773 -0.5600 0.4168 -0.2733
0.7945 1.0603 1.5393 0.0098
I have the above matrix and I'd like to perform Gauss elimination on it with MatLab such that I am left with an upper triangular matrix. Please how can I proceed?

Antworten (3)

József Szabó
József Szabó am 29 Jan. 2020
function x = solveGauss(A,b)
s = length(A);
for j = 1:(s-1)
for i = s:-1:j+1
m = A(i,j)/A(j,j);
A(i,:) = A(i,:) - m*A(j,:);
b(i) = b(i) - m*b(j);
end
end
x = zeros(s,1);
x(s) = b(s)/A(s,s);
for i = s-1:-1:1
sum = 0;
for j = s:-1:i+1
sum = sum + A(i,j)*x(j);
end
x(i) = (b(i)- sum)/A(i,i);
end
  4 Kommentare
Brinzan
Brinzan am 25 Nov. 2024
There are no input arguments, what do I do, qnd I dont even know where to put or how much to put like at the Matrix for A do I just write A=[00011110011],[0100001111],[0111110000] or what în the Code cuz it just not working.
Walter Roberson
Walter Roberson am 25 Nov. 2024
A = randi([-1 2], 5, 5)
A = 5×5
2 -1 0 0 2 0 -1 -1 2 1 1 2 2 2 0 -1 0 0 -1 2 -1 0 1 2 -1
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b = randi([-2 2], 5, 1)
b = 5×1
1 1 -2 -1 -2
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x = solveGauss(A, b)
x = 5×1
0.7059 0.0588 -1.3529 -0.0588 -0.1765
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function x = solveGauss(A,b)
s = length(A);
for j = 1:(s-1)
for i = s:-1:j+1
m = A(i,j)/A(j,j);
A(i,:) = A(i,:) - m*A(j,:);
b(i) = b(i) - m*b(j);
end
end
x = zeros(s,1);
x(s) = b(s)/A(s,s);
for i = s-1:-1:1
sum = 0;
for j = s:-1:i+1
sum = sum + A(i,j)*x(j);
end
x(i) = (b(i)- sum)/A(i,i);
end
end

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Richard Brown
Richard Brown am 12 Jul. 2012
The function you want is LU
[L, U] = lu(K);
The upper triangular matrix resulting from Gaussian elimination with partial pivoting is U. L is a permuted lower triangular matrix. If you're using it to solve equations K*x = b, then you can do
x = U \ (L \ b);
or if you only have one right hand side, you can save a bit of effort and let MATLAB do it:
x = K \ b;
  2 Kommentare
Lukumon Kazeem
Lukumon Kazeem am 12 Jul. 2012
Thank you Richard for your response. I have used this approach a no. of times ([L U]=lu(k)) and the results are always different from that in published literature. I suspect it's because it performs partial and not complete pivoting
Richard Brown
Richard Brown am 13 Jul. 2012
I wouldn't expect it would necessarily compare with published literature - what you get depends on the pivoting strategy (as you point out).
Complete pivoting is rarely used - it is pretty universally recognised that there is no practical advantage to using it over partial pivoting, and there is significantly more implementation overhead. So I would question whether results you've found in the literature use complete pivoting, unless it was a paper studying pivoting strategies.
What you might want is the LU factorisation with no pivoting. You can trick lu into providing this by using the sparse version of the algorithm with a pivoting threshold of zero:
[L, U] = lu(sparse(K),0);
% L = full(L); U = full(U); %optionally

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James Tursa
James Tursa am 11 Jul. 2012
  2 Kommentare
Lukumon Kazeem
Lukumon Kazeem am 12 Jul. 2012
James, thank you for your response. I have tried to apply your suggestion to the matrix I posed earlier but it came up with the below prompt. What do you think?
"??? Undefined function or method 'gecp' for input arguments of type 'double'".
James Tursa
James Tursa am 13 Jul. 2012
You need to download the gecp function from the FEX link I posted above, and then put the file gecp.m somewhere on the MATLAB path.

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