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Leading on from a question on the practicality of inverse matrices used in physical engineering systems which I've asked previously, I have come across an issue with my system where I want to take the inverse of a square Matrix (see below) to solve for a set of other variables but I cannot due to its determinant equalling zero.

In my simple system where $W = F^{-1} {\cdot} R$

Below is the matrix $F$ which I'm having trouble getting an inverse for, so I can't seem to solve my system. System Matrix Equation

From this, I have the following questions:

  1. Why do determinants equalling zero dictate non-invertible (or "singular"/"degenerate") matrices?
  2. In the cases where variables I've inputted into my system lead to determinants equalling zero, is there a work around to still somehow get the inverse of this matrix or otherwise solve the system of equations I'm wanting? (i.e. rearranging rows and columns or using some other fancy matrix identities or mathematics?
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    $\begingroup$ When your matrix is singular, the corresponding system of linear equations can have no solutions or many solutions. But try looking up the method of least squares and the Moore-Penrose inverse. $\endgroup$
    – nasekatnasushi
    Dec 2, 2022 at 9:43
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    $\begingroup$ For the physical reality , the condition of a matrix matters. If a matrix has low condition , it is numerically stable , variating the entries does not influence the determinant much (this corresponds with stable systems). If the condition is high , variating the entries influences the determinant very strong (it can change its sign, become $0$ or grow drastically. This corresponds with instable systems) $\endgroup$
    – Peter
    Dec 2, 2022 at 10:45
  • $\begingroup$ Great comments thank you! @nasekatnasushi in the case of many solutions how does the Moore-Penrose inverse with least squares method discriminate between solutions? Is the "best" solution simply the one with the best least squares fit which is automatically given by the Moore-Penrose inverse? $\endgroup$
    – Hendrix13
    Dec 2, 2022 at 11:52
  • $\begingroup$ @Peter, Thank you for the practical context, would it be better to get the condition of my final solution matrix or the inverse / pseudoinverse matrices to measure their stability? $\endgroup$
    – Hendrix13
    Dec 2, 2022 at 11:53
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    $\begingroup$ @Hendrix13 You get a solution with minimal Euclidean norm. I am a little rusty on this, so I would just be quoting Wikipedia: en.wikipedia.org/wiki/… $\endgroup$
    – nasekatnasushi
    Dec 2, 2022 at 12:33

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