Question

im ridiculous...i wrote this 3-4 years ago and now i cant figure it out !!!! this is for solving n linear equations with n unknowns

Answer 1

function G =GJ(a,b)
w=length(a);
if size(a,1)~=size(a,2)
disp('This is not a n*n system!!!');
else
for j=1:w
for i=1:w
if a(i,j)~=0
l(j)=i;
end
end
end
end
c=[a b];
for m=1:w
cm=c(m,:);
c(m,:)=c(l(m),:);
c(l(m),:)=cm;
for n=1:w
if l(n)==m
l(n)=l(m);
end
end
end
for i=1:w-1
for j=i+1:w
cj=c(j,:);
ci=c(i,:);
cj=cj-(c(j,i)/c(i,i))*ci;
c(j,:)=cj;
end
end
for i=w:-1:2
for j=i-1:-1:1
cj=c(j,:);
ci=c(i,:);
cj=cj-(c(j,i)/c(i,i))*ci;
c(j,:)=cj;
end
end
for i=1:w
x(i)=c(i,w+1)/c(i,i);
end
c
fprintf('%f\n',x);

Answer 2

@experimentX @phi

Answer 3

any better ideas for that

Answer 4

is it in matlab?

Answer 5

yes Gauss-Jordan method

Answer 6

Hold on ... let me run!!

Answer 7

what type of input does function take?

Answer 8

a=n*n matrix
b=1*n matrix

Answer 9

hold on man its not working :) i thought its right

Answer 10

do you put symbolic values for b?

Answer 11

OH .. hell. I've been so forgetful.

Answer 12

Shouldn't this be easy ... since matlab stands for Matrix Laboratory still you managed to code it all. You should have done it in C perhaps ... lol. just joking.

Answer 13

lol

Answer 14

you might be interested in this
http://projecteuler.net/

Answer 15

I assume you know
\ Backslash or left matrix divide.
A\B is the matrix division of A into B, which is roughly the
same as INV(A)*B , except it is computed in a different way.

If A is an N-by-N matrix and B is a column vector with N
components, or a matrix with several such columns, then
X = A\B is the solution to the equation A*X = B computed by
Gaussian elimination. A warning message is printed if A is
badly scaled or nearly singular. A\EYE(SIZE(A)) produces the
inverse of A.

Just for giggles:

% gaussian elimination (not sophisticated)
% ASSUMES square N x N matrix
% tries not to divide by 0, but does not swap rows

% create a 3x3 and a 3x1
a= rand(3); b= rand(3,1);

c= [a b] % create the augmented matrix
N= length(a);
for ii=2:N
jj= ii-1;
% the pivot is c(jj,jj) on the diagonal c(1,1) is the 1st pivot
% the next statement creates a submatrix (using an outer product) by
% the key position (the position that will be zeroed out) for each
% row. All positions below the pivot will be zeroed out by the
% following statement.
if (abs(c(jj,jj))>1e-12) % if non-zero use this pivot
x= c(ii:end,jj)/c(jj,jj) * c(jj,jj:end);
c(ii:end,jj:end)= c(ii:end,jj:end) - x;
end;
end;

% Now work backwards. First normalize the pivot position to 1
% Then zero out all entries above the pivot position
for jj=N:-1:1
ii= jj-1;
if (abs(c(jj,jj))>1e-12) % if non-zero use this pivot
c(jj,jj:end)= c(jj,jj:end)/c(jj,jj);
x= c(1:ii,jj) * c(jj,jj:end);
c(1:ii,jj:end)= c(1:ii,jj:end) - x;
end;
end;

c(:,N+1:end)
a\b % compare to Matlab's gaussian