Ax=b solution with singular matrix.

%x+2y+3z=14

%4x+5y+6z=32

%7x+8y+9z=41

(actually, x=1,y=2,z=3)

A=[1 2 3;4 5 6;7 8 9]

b=[14;32;41]

x=A\b

Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 1.093272e-16.

%in order to eliminate this singularity singular value decomposition strongly recommended.

How can I perform svd for solving this equation properly?

Answer 1

[Matt J]

 pinv(A)*b

[sermet]

it gives

ans =

-1.749999999999991

   1.500000000000001

   4.749999999999994

but these are not the roots,

the roots are

1

2

3

%which are x y z

[Roger Stafford]

No, 1,2,3 are not roots. They don't satisfy the third equation. Your three equations are actually mutually incompatible. They have no roots at all. Subtract the first equation from the second to get:

 3*x + 3*y + 3*z = 18

Subtract the second equation from the third:

 3*x + 3*y + 3*z = 9

No numbers could possibly satisfy this, so there can be no solution.

[sermet]

ok, Roger, thank you I just made mistake.

[Roger Stafford]

A geometrical interpretation of these equations would be that they each represent a plane in 3D. Two of the planes intersect in a line, and the third plane is parallel to this line, but never intersects it, so the three planes never have a common intersection. Even trying to find a "best solution" has no unique answer - all points along a certain infinite line would have the same least squares error.