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Impossible Ax=b is solved by linesolve(A,b) and mldivide(A,B)

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A=[-1 2 3;0 1 -1;2 0 5;1 3 2]
B=[1 1 1 1]'
x=linsolve(A,B)
x =
0.1111
0.3333
0.1111
x=mldivide(A,B)
x =
0.1111
0.3333
0.1111
HOwever A*x
ans =
0.8889
0.2222
0.7778
1.3333
and b= [1 1 1 1]'
If I solve the system in paper, the system is impossible but matlab says it´s possible. When I test matlab answer, it is wrong. If I do A=sym([-1 2 3;0 1 -1;2 0 5;1 3 2])
colspace(A) A =
[ -1, 2, 3] [ 0, 1, -1] [ 2, 0, 5] [ 1, 3, 2]
ans =
[ 1, 0, 0] [ 0, 1, 0] [ 0, 0, 1] [ 1/3, 7/3, 2/3]
The system seems possible but I think it isn't. Can someone tell me what's going on? How do I know if the system is possible?

Accepted Answer

Matt Fig
Matt Fig on 14 Aug 2012
Edited: Matt Fig on 14 Aug 2012
From the documentation:
" If A is an m-by-n matrix with m ~= n and B is a column vector with m components, or a matrix with several such columns, then X = A\B is the solution in the least squares sense to the under- or overdetermined system of equations AX = B. In other words, X minimizes norm(A*X - B), the length of the vector AX - B. The rank k of A is determined from the QR decomposition with column pivoting. The computed solution X has at most k nonzero elements per column. If k < n, this is usually not the same solution as x = pinv(A)*B, which returns a least squares solution. "
You have 4 equations and 3 unknowns. This is an overdetermined system, so MATLAB is following the documentation.
  5 Comments
Matt Fig
Matt Fig on 14 Aug 2012
Sure, just make a little function.
isexact = @(A,x,B) max((A*x-B))<1e-13; % Choose your tol.
Now,
A=[-1 2 3;0 1 -1;2 0 5;1 3 2];
B=[1 1 1 1]';
x = A\B;
isexact(A,x,B) % Returns 0 == not exact.
But, let's try with:
A=[-1 2 3;0 1 -1;2 0 5];
B=[2 3 4]';
x = A\B;
isexact(A,x,B) % Returns 1 == yes exact
Pedro
Pedro on 14 Aug 2012
wow.. didn't know matlab was that easy to program. Thanks!

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