incorrect result in solving system using MATLAB ?

I am trynig to solve the following system (mat=coefficient matrix,B=output matrix)

mat =
       1     2     6
       2     5    14
       5     7    24
B=
   0
   0
   0

the expected result is not zeros for sure because this system is linear dependent . And I used mat/B and linsolve(mat,B).However , I have got the answer

ans=
       0 
       0 
       0

and this warning

Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND =
1.009294e-18.

could any one help me why I got this result and what I should to get the correct answer?

Answer 1

[Torsten]

Try

Z = null(mat)

It should give you a vector Z different from the zero vector that solves

mat*Z = 0

Best wishes

Torsten.

[Eliza]

I have got this answer

1.0e-14 *
  0.0444
       0
  0.1776

[Eliza]

The complete question of this is system: We have nontrivial solution of [b1 b2 b3]=[-12 -12 6].This implies the three given vectors are linearly dependent .Can you find another nontrivial solution ?

[Torsten]

1.0e-14 *
  0.0444
       0
  0.1776

is what you get when you run

mat = [1     2     6 ;  2     5    14 ; 5     7    24]
Z = null(mat);
Z

?

Then your MATLAB version has a bug.

Result should be

[-2/3 -2/3 1/3] or [2/3 2/3 -1/3]

Best wishes

Torsten.

[Eliza]

No this result of this operation mat*Z to check if it gives the 0 or not

[Torsten]

And what do you get for Z ?

Because this is one of the vectors you've been searching for, I thought ?

Best wishes

Torsten.

[Walter Roberson]

Any non-zero scalar multiple of [-12; -12; 6] is also a solution. [-12;-12;6] is 18 * null(mat)

[Eliza]

I have got Z=-0.6667 -0.6667 0.3333 . However,I was having some problem in understanding null space. I was expect the result as I did it manually .So thanks all.

[Steven Lord]

Was the result of null(mat, 'r') closer to what you found when you did it manually? From the help for the null function:

"The orthonormal basis is preferable numerically, while the rational basis may be preferable pedagogically."

The rational basis is what you get when you add the 'r' flag.

Answer 2

[Steven Lord]

That answer is one solution to your system of equations. It is not the only solution to that system.

If you add x and any linear combination of columns from the nullspace of your matrix that will be another solution to your system. Take a look at the null function for more information.

Answer 3

[Walter Roberson]

rank(mat) is 2 because the third column is twice the sum of the other two columns. You cannot use the \ operator with singular matrix.

If the b matrix was not all zero then pinv(mat)*b might work, but since it is all 0 the result is going to be all 0.

With the all-zero b, you are effectively looking null spaces, for which you should look at null(mat)

[Eliza]

My goal of solving this system is to prove that it is not linear independent ,so I make it equal to b matrix which is equal to zero and if I get result rather than zeros this means it is Linear dependent ,but the result gave me zeros .All what I want now is getting any result except zeros .

[Walter Roberson]

11 * [1 2 6] - 3 * [2 5 14] = [5 7 24] . Therefore the last row does not add any new information to the matrix and the rank is 2.

"and if I get result rather than zeros this means it is Linear dependent"

No, it would mean that they were linearly independent