incorrect result in solving system using MATLAB ?

2017 11 14
3 个回答
回答已采纳
更新时间：2017 11 16
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0 个投票

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?

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 采纳的回答

Torsten
Torsten 2017-11-15

0 个投票

Try
Z = null(mat)
It should give you a vector Z different from the zero vector that solves
mat*Z = 0
Best wishes
Torsten.

8 个评论
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Eliza
Eliza 2017-11-15
I have got this answer
1.0e-14 *
0.0444
0
0.1776
Eliza
Eliza 2017-11-15
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
Torsten 2017-11-15
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
Eliza 2017-11-15
No this result of this operation mat*Z to check if it gives the 0 or not
Torsten
Torsten 2017-11-15
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
Walter Roberson 2017-11-15
Any non-zero scalar multiple of [-12; -12; 6] is also a solution. [-12;-12;6] is 18 * null(mat)
Eliza
Eliza 2017-11-16
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
Steven Lord 2017-11-16
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.

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更多回答（2 个）

Steven Lord
Steven Lord 2017-11-14

0 个投票

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.
Walter Roberson
Walter Roberson 2017-11-14
编辑：Walter Roberson 2017-11-14

0 个投票

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)

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Eliza
Eliza 2017-11-14
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
Walter Roberson 2017-11-14
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

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