incorrect result in solving system using MATLAB ?
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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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채택된 답변
Torsten
2017년 11월 15일
Try
Z = null(mat)
It should give you a vector Z different from the zero vector that solves
mat*Z = 0
Best wishes
Torsten.
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Steven Lord
2017년 11월 16일
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.
추가 답변 (2개)
Steven Lord
2017년 11월 14일
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.
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Walter Roberson
2017년 11월 14일
편집: Walter Roberson
2017년 11월 14일
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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Walter Roberson
2017년 11월 14일
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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