Why linsolve cannot solve this very simple equation?

A = [4,2,2; 5,1,3; 6,0,4];
B = [60; 70; 80];
X = linsolve(A,B)

The solution should be 6, 7, 11, since:

4x6 + 2x7 + 2x11 = 60

5x6 + 1x7 + 3x11 = 70

6x6 + 0x7 + 4x11 = 80

But I get the following answer:

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

7.401487e-18.

X = [0; 10; 20]

Why?

Answer 1

[Stephan]

To solve this, the rank should be 3. Row 1 and 3 are not linear independent.

A = [4,2,2; 5,1,3; 6,0,4]
B = [60; 70; 80];

r1 = rank(A)
r2 = rank([A, B])

linsolve gives a correct solution, because there are more then 1 solutions, due to the rank:

>> X = linsolve(A,B)
Test = A*X
Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND =
7.401487e-18. 
 

X =

     0
    10
    20


Test =

    60
    70
    80

Answer 2

[Steven Lord]

The vector [6; 7; 11] is a solution to the problem but it is not the only solution.

A = [4,2,2; 5,1,3; 6,0,4];
B = [60; 70; 80];

sol1 = [6; 7; 11];
check = A*sol1-B; % should be close to 0
[sol1, check]

ans = 3×2

6 0 7 0 11 0

N = null(A); % A*N is close to the 0 vector
sol2 = sol1 + N; % Since A*sol1 = B and A*N = 0, A*(sol1+N) = B+0 = B

check2 = A*sol2-B; % should also be close to 0
[sol2, check2]

ans = 3×2

5.4655 0 7.2673 0 11.8018 0

sol3 = sol1 + 42*N; % A*(sol1+42*N) = A*sol1 + 42*A*N = B+0 = B
[sol3, A*sol3-B] % Also close to 0

ans = 3×2

-16.4499 -0.0000 18.2250 -0.0000 44.6749 0.0000