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← Matrix2.5Determinants2

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Showing Revision 1 created 02/25/2021 by jnelson42.

  1. Let's now compute the determinant using
    the previous formula of a 3 by 3 matrix.
  2. So there's my matrix A, let's say.
  3. And, once again, I'm allowed to expand
    along any row or column and I'll get
  4. the equivalent result when I compute
    this determinant.
  5. And then in conclusion,
    depending on whether that number is
  6. zero or not, that tells me whether or not
    the original matrix was invertible.
  7. I'm just going to use just that first
    general formula we wrote here
  8. for a determinant by expanding along
    the first row.
  9. So I take the number 5 and then I
    multiply by the
  10. determinant of the corresponding submatrix
    which is now 0, 2, -1, 3. And then I go
  11. minus, I alternate signs, so check this
    minus -3, goes +3 so we flip that and
  12. then I remove again row 1 column 2,
    and I just plug in the determinant here,
  13. 1,2,2,3, the corresponding submatrix.
  14. And then lastly I pluck out the 2, the
    last entry in that row, and then I
  15. multiply that by the corresponding
    submatrix here, that
  16. determinant here 1, 0, 2, -1.
    So there's my determinant formula.
  17. And again, just as a friendly reminder,
    the determinant of a 2x2 matrix is this
  18. sort of cross multiplication process,
  19. So let's go ahead and clear this up here.
  20. So we have 5, ad-bc, so this is 0-(-2) so
    this is going to be a +2, so 5(2).
  21. So plus 3 times ad, which is
    3-(-4) is -1.
  22. And lastly we have +2, ad-bc, is -1.
  23. So we have that here. And let's add
    these numbers up.
  24. We get 10-3-2 and that results in 5.
  25. That means that my original matrix A
    here, is invertible.
  26. So there's some matrix out there I
    could find and multiply by that inverse
  27. matrix and then produce the identity.
  28. Since we haven't done a column yet,
    let's expand along column 2 here.
  29. And we will compute the determinant
    that way and we see that we also get 5,
  30. but we'll have to do little less
  31. work because of the 0 here.
  32. I just want to note right we have this
    checkerboard pattern with the signs.
  33. I'm just going to remind you for
    the determinant we have a negative with
  34. that term, a positive and then a negative,
    we alternate signs.
  35. Let's now compute the determinant of A
    one more time, just expanding along this
  36. time column 2. Ok, so I have, negative -3
    which is +3 times the determinant of
  37. the submatrix, the same as it was before.
  38. So 1, 2, 2, 3. Ok, now alternate sign +0,
    ok I'll just leave this as a place holder
  39. so we can see it. Plus zero. Of course,
    that'll go away.
  40. But the submatrix associated with zero
    when I remove now column 2 and row 2 is
  41. going to be 5, 2, 2, 3. Ok so that just
    goes away. And then I go minus -1
  42. which is plus 1 when I alternate sign,
    and then I multiply by the determinant
  43. of the submatrix corresponding with the
    -1 here, so I'm going to remove row 3
  44. and column 2 and that leaves me
    with 5, 2 and 1, 2.
  45. Let's compute that determinant.
    So again, I use the formula
  46. for a 2x2 determinant ad-bc here. Right,
    so 3 times 1(3),ad, 3-4 is -1. Ok. plus 0,
  47. and then plus 1 times 10-2 is going to be
    8. So what I have when the dust settles
  48. is -3+8 sure enough results in 5.
    Point being the determinant can be
  49. ascertained by expanding along any
    row or column we get a consistent result.