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Problem using Left Hand Division in MATLAB

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Michael on 16 Oct 2014
Commented: Michael on 16 Oct 2014
Hello Everyone,
I am currently working on trying to solve a System of Linear Equations, but MATLAB keeps on returning all zeroes. This is because all of the elements in the second matrix (on the answer side of the linear equations) are zeroes, but that is the only thing I am allowed to work with.
Here is the code:
sysEq = [1/sqrt(35), -3/sqrt(34), 1/sqrt(42); 3/sqrt(35), 0, -4/sqrt(42); 5/sqrt(35), 5/sqrt(34), 5/sqrt(42)]
sysEqAns = [0, 0, 0]';
Ans = sysEq\sysEqAns
If anyone knows how to resolve this complication, I'd be very grateful.
Thanks, Mike


Guillaume on 16 Oct 2014
Well, [0 0 0]' is obviously the solution to your particular set of linear equations, so I'm not sure what else you're hoping to get.
I don't particularly understand your comment 'that is the only thing I am allowed to work with'. If your constants are not zero, obviously don't set them as 0, otherwise you'll never get the right answer.
Finally, as an aside, I wouldn't call my variable Ans, it's a bit too close to matlab built-in variable ans.
Mischa Kim
Mischa Kim on 16 Oct 2014
Not that obvious. For a rank-deficient matrix there are an inifinte number of non-zero solution vectors.

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Mischa Kim
Mischa Kim on 16 Oct 2014
Edited: Mischa Kim on 16 Oct 2014
Michael, the answer (zero-vector) is correct. If you do
you'll see that the matrix has full rank (= 3). This means that the rows (and columns) of the matrix are linearly independent which in turn means that only Ans = [0 0 0]' multiplied by the matrix will result in the zero-vector sysEqAns = [0 0 0]'. In other words, there is no other solution for Ans that results in the zero-vector for sysEqAns.

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