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Solving for the Determinent of a Matrix

  1. Aug 5, 2004 #1
    Okay, I'm learning currently how to solve for the determinent of a Matrix. Of course the book explains how to solve for a 2 X 2 Matrix, a 3 X 3 Matrix, a 4 X 4 Matrix, ect. But it says nothing about how to solve for a 3 X 2 Matrix.

    Any idea how to do this? I'm really baffled on this.
  2. jcsd
  3. Aug 6, 2004 #2
    Just in case no one knows what I'm talking about, by a 3 X 2 Matrix, I mean like this:

    [1 0 -2]
    [2 1 -1]

    It's very hard to figure this out. My book doesn't even go over it. They'll go over 2 X 2, 3 X 3, 4 X 4, 5 X 5, 6 X 6, ect.. But for some reason they don't touch on if the Matrix doesn't have equal sides.
  4. Aug 6, 2004 #3
    No wonder, the determinant function (or "a determinantal function") is defined as a function from the set of all nxn (i.e square) matrices (with elements in a field F), to the field F (the determinant takes a square matrix and spits back out a number). There are reasons for this.

    "The" inverse of a matrix A is a matrix B such that AB = BA = I (i.e. B is both a left inverse and a right inverse of A). Suppose A is of size nxm, and B is pxq. Then AB is nxq and BA is pxm. But I is square, say I is a txt matrix. Since AB = BA = I, this forces n = q = p = m = t, i.e. A and B are both square. Thus only square matrices can have inverses.

    One wants the determinant function to characterize when exactly a matrix X has an inverse (it just so happens to be that X has an inverse iff det(X) != 0). But according to the above, only square matrices can have inverses, so it doesn't make much sense to define the determinant for non-square matrices.

    Btw, a 3x2 matrix has 3 rows and 2 columns. The matrix you used as an example is a 2x3 matrix.
    Last edited: Aug 6, 2004
  5. Aug 6, 2004 #4
    So is it possible to solve the determinent of non square Matrices? Or is the answer simply Cannot be calculated?
  6. Aug 6, 2004 #5
    No, using the ordinary definition of the determinant function, you cannot calculate the determinant of a non-square matrix.
  7. Aug 6, 2004 #6


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    It's not so much that you "cannot calculate" it as that it is not defined!

    The determinant is only defined for square matrices.

    Asking how to find the determinant of a 3 by 2 matrix is a lot like asking how to find the square root of a chair.
  8. Jul 1, 2007 #7
    you can solve a 2x3 matrix, you use this all the time to when multiplying vectors.

    [a b c]
    [e f g]

  9. Jul 2, 2007 #8


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    So, this is called the korciuch-operator? :biggrin:
  10. Jul 2, 2007 #9


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    The phrase "solve a matrix" doesn't make sense. The orginal question was "solve for the determinant of a matrix". Again, a 3 by 2 matrix (or any non-square matrix) does not have a determinant.
  11. Jul 2, 2007 #10
    thats the cross product, which in this case, is a 3x3 matrix with [i j k] as the first row:
    [i j k]
    [a b c]
    [e f g]
  12. Jul 4, 2007 #11


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    the det of a non square matrix is either not defined or zero.
  13. Jul 6, 2007 #12


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    Could you give an example of a non-square matrix that has determanant 0?
  14. Jul 8, 2007 #13
    You could try the products of the nonzero singular values in the singular value decomposition. It's not the determinant, but it's the closest thing you're going to get.
  15. Jul 8, 2007 #14


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    I was wondering why mathwonk said
  16. Jul 8, 2007 #15
    I was wondering the same thing too, when I came across his post. I don't think the guru would have made a mistake over something like this. So I keep wondering......

    Edit: Googling does throw up some links about non square determinants. I still can't understand the motivation to assign a 0 value to a non square det.
    Last edited: Jul 8, 2007
  17. Jul 9, 2007 #16
    mathwonk is right. You can't get the determinant of a non square matrix. The determinant of a matrix is the n dimensional volume spanned by n vectors. If you don't have n vectors this value either can't be obtained, or is zero.

    I believe there is a way of asking for the k<n dimensional volume of k, n-dimensional, vectors, but I don't know how one would go about this. You could try a change of coordinates , via the singular value decomposition say. I believe the products of the non zero singular values do give you the k dimensional volume spanned by k column vectors. I don't have any references for this, but I'm pretty sure it's ok.
  18. Jul 9, 2007 #17


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    of course you can compute all the determinants of square sub matrices.

    e.g to coordinatize planes in 4 space, you can consider them as sappned by the columns of 4x2 matrices, and compute all, lets see, 4 choose 2 is 6, 2x2 subdeterminants.

    this gives projective coordinates on the grassmannian of 2 plkanes in 4 space, embedding it in projective 5 space, since changing the basis for a plane changes those determinants all by the same scalar factor.

    i.e. essentially yiu are forming a parallelogram in your plane, then taking all 6 projections of that parallelogram onto pairs of axes, and computing their oriented areas.

    by the pythagorean theorem for areas, this also allows computation of the area of the original parallelogram in 4 space.
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