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

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?

Answer 1

[Matt Fig]

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.

[Matt Fig]

Comment by Pedro:

"Thanks for the answer but what i really wanted was to ask matlab if the system is possible or impossible, instead of doing all those aproximations. How do I ask that?"

[Matt Fig]

Your question is a little vague. By "impossible" do you mean not exact? Have a look at the RANK and COND functions. If a matrix is non-square you are most likely into least-squares territory where a fit will be made.

[Pedro]

yes by impossible I mean not exact. Seems like you just have to check the answer like I did, multipling A*x and see if it is equal to b... right?

[Matt Fig]

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]

wow.. didn't know matlab was that easy to program. Thanks!