Notices
Results 1 to 11 of 11

Thread: 2x2 matrix determinant

  1. #1 2x2 matrix determinant 
    Forum Sophomore
    Join Date
    Jan 2008
    Posts
    130
    Take the following:

    • M = |b1,b2|


    where M is a matrix and b1 and b2 are the basis vectors for the coordinate frame. b1 and b2 are defined as:

    • b1 = (a,b)
      b2 = (c,d)


    so our matrix looks like:

    • |a,c|
      |b,d|


    i.e. column major matrix.

    The determinant of a scalar is simply the scalar itself. This makes sense as the 'area' of a scalar would be itself (please correct me if I sound delusional) - just using the concept of area applied to a scalar to provide a concept for the determinant, which ultimately returns an area right?

    The determinant of a 2x2 matrix M is:

    • |M| = a*d - c*b


    abs(|M|) is equal to the area of the parallelogram bounded by the basis vectors b1 and b2.

    I'm struggling to see the relationship of the determinant formula, i.e. how does the x component of the first basis vector multiplied by the y component of the second basis vector have significance?

    Currently I'm grasping the idea that the determinant forumula is done this way to compute the signed area of the parallelogram, thus 'preserving' the 'handedness' of the coordinate frame, but beyond that I'm lost...

    I tried computing the area of the parallelogram bounded by the two basis vectors, as:

    • A = ||b1||||b2||


    where ||x|| is the magnitude of vector x.

    This should work fine, because the area of a parallelogram is the product of its adjacent sides. However I can't find anything relating the determinant to this calculation.


    Reply With Quote  
     

  2.  
     

  3. #2 Re: 2x2 matrix determinant 
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by rgba
    Take the following:

    • M = |b1,b2|


    where M is a matrix and b1 and b2 are the basis vectors for the coordinate frame. b1 and b2 are defined as:

    • b1 = (a,b)
      b2 = (c,d)


    so our matrix looks like:

    • |a,c|
      |b,d|


    i.e. column major matrix.

    The determinant of a scalar is simply the scalar itself. This makes sense as the 'area' of a scalar would be itself (please correct me if I sound delusional) - just using the concept of area applied to a scalar to provide a concept for the determinant, which ultimately returns an area right?

    The determinant of a 2x2 matrix M is:

    • |M| = a*d - c*b


    abs(|M|) is equal to the area of the parallelogram bounded by the basis vectors b1 and b2.

    I'm struggling to see the relationship of the determinant formula, i.e. how does the x component of the first basis vector multiplied by the y component of the second basis vector have significance?

    Currently I'm grasping the idea that the determinant forumula is done this way to compute the signed area of the parallelogram, thus 'preserving' the 'handedness' of the coordinate frame, but beyond that I'm lost...

    I tried computing the area of the parallelogram bounded by the two basis vectors, as:

    • A = ||b1||||b2||


    where ||x|| is the magnitude of vector x.

    This should work fine, because the area of a parallelogram is the product of its adjacent sides. However I can't find anything relating the determinant to this calculation.
    I think that a good deal of your problem is that A = ||b1||||b2|| is only true for a rectangle and not for a general parallelogram.


    Reply With Quote  
     

  4. #3  
    Forum Sophomore
    Join Date
    Jan 2008
    Posts
    130
    Cool, I'll look into that again! Thanks!
    Reply With Quote  
     

  5. #4  
    Forum Sophomore
    Join Date
    Jan 2008
    Posts
    130
    Hey, well just had a look. I made a mistake, its actually:

    • area = ||b1||||b2||sinTHETA


    Where b2 is the slant vector, b1 is the base and THETA is the angle between the two vectors.

    I still can't see the relationship between the following:

    • |M| = ||b1||||b2||sinTHETA
      a*d - b*c = sqrt(a*a + b*b) * sqrt(c*c + d*d) * sin THETA


    Am I correct in using the assumption that the determinant of the basis vectors (i.e. the ones defining the parallelogram) is the same as using ||b1||||b2||sinTHETA, just with a possible sign difference?

    Is it related to finding the angle between the two vectors, i.e. dot product?

    Hope you can help out! Thanks!
    Reply With Quote  
     

  6. #5  
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by rgba
    Hey, well just had a look. I made a mistake, its actually:

    • area = ||b1||||b2||sinTHETA


    Where b2 is the slant vector, b1 is the base and THETA is the angle between the two vectors.

    I still can't see the relationship between the following:

    • |M| = ||b1||||b2||sinTHETA
      a*d - b*c = sqrt(a*a + b*b) * sqrt(c*c + d*d) * sin THETA


    Am I correct in using the assumption that the determinant of the basis vectors (i.e. the ones defining the parallelogram) is the same as using ||b1||||b2||sinTHETA, just with a possible sign difference?

    Is it related to finding the angle between the two vectors, i.e. dot product?

    Hope you can help out! Thanks!
    It is not closely related to the dot product, although the dot product is at the heart of the geometry.

    One way to get at this problem is to look at the special case in which b=0, so that your first vector lies along the x-axis. Apply the formula for area of a parallelogram Area = base x height to get immediately that Area = a*d. But that is the determinant that you are considering in the special case in which b = 0.

    For the general case you can then apply a rigid rotation to move rotate both (a,b) and (c,d) so that the transformed (a,b) lies along the x-axis. Since the rigid rotation preserves area of the parallelogram you know that the area is the determinant of the matrix whose columns are the transformed vectors. Then take that determinant and apply some simple algebra and the trig identity that cos^2 +sin^2 = 1 and you will find that what you have is the determinant that you seek.
    Reply With Quote  
     

  7. #6  
    Forum Sophomore
    Join Date
    Jan 2008
    Posts
    130
    Hey DrRocket,

    Thanks for the reply, I'll come back to you as soon as I get a chance! Been hectic busy, have to go to a meeting now, then once that's done I'll be back on the math!

    Really appreciate your help!
    Reply With Quote  
     

  8. #7  
    Forum Sophomore
    Join Date
    Jan 2008
    Posts
    130
    Hey, finally got a chance to sit down and read your post properly! Well, it's surprisingly intuitive after all!

    Taking this as an example:

    • b1 = (2,1)
      b2 = (-2,2)

      matrix is:
      |2 -2|
      |1 2|


    It seems you can effectively discard opposite components both ways, get the product of their lengths and sum to get the determinant, for example:

    • b1a = (2,0)
      b2a = (0,2)

      b1a_len = 2
      b2a_len = 2

      b1b = (0,1)
      b2b = (-2,0)

      b1b_len = 1
      b2b_len = 2

      area = b1a_len*b2a_len + b1b_len*b2b_len
      = 2*2 + 1*2
      = 6

      determinant = b1x*b2y - b1y*b2x
      = 2*2 -(1*-2)
      = 6


    That's pretty cool I suppose this works because the vectors have a relationship with each other? Can we say that we break the vectors down like above (determinant forumla) to define each vector on a single axis, and then sum that for each component discarded?

    I'm not quite sure what you mean by this:

    Then take that determinant and apply some simple algebra and the trig identity that cos^2 +sin^2 = 1 and you will find that what you have is the determinant that you seek.
    Sorry for being so slow! Thanks for the great help!
    Reply With Quote  
     

  9. #8  
    Forum Sophomore
    Join Date
    Jan 2008
    Posts
    130
    Also, this came to my mind, it's basically subtracting the component areas right?
    Reply With Quote  
     

  10. #9  
    Forum Sophomore
    Join Date
    Jan 2008
    Posts
    130
    Hey,

    I've been looking at the determinant of a 3x3 matrix, and it appears to be identical to the cross product of two vectors, only that the determinant returns a scalar and the cross product returns a vector.





    Taking as an example:




    It appears that each component of the cross product is the area of the plane perpendicular to itself (i.e. the 2x2 determinant of the other vectors), for example, the component is equal to the area of the parallogram bounded by the and vectors projected onto the plane, is this correct, and why so?

    I'm finding this hard to understand as I can't quite relate the areas applied to each component giving a vector perpendicular to the plane bounded by the and vectors!

    Hope you can help out! I feel I'm making a little progress on this front, thanks to the help I've been getting here

    BTW: This latex support is pretty cool, but takes some getting used to
    Reply With Quote  
     

  11. #10  
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by rgba
    Hey,

    I've been looking at the determinant of a 3x3 matrix, and it appears to be identical to the cross product of two vectors, only that the determinant returns a scalar and the cross product returns a vector.





    Taking as an example:




    It appears that each component of the cross product is the area of the plane perpendicular to itself (i.e. the 2x2 determinant of the other vectors), for example, the component is equal to the area of the parallogram bounded by the and vectors projected onto the plane, is this correct, and why so?

    I'm finding this hard to understand as I can't quite relate the areas applied to each component giving a vector perpendicular to the plane bounded by the and vectors!

    Hope you can help out! I feel I'm making a little progress on this front, thanks to the help I've been getting here

    BTW: This latex support is pretty cool, but takes some getting used to
    In the 3x3 case if you form a matrix taking as rows vectors C,B,A then the determinant is (A X B). C and it is fairly easy to see that the determinant is up to sign the volume of parallelopiped formed by the three vectors. It is not hard to prove this to yourself, but if you want to see it done with pictures you might take a look at Advanced Calculus by Kaplan.
    Reply With Quote  
     

  12. #11  
    Forum Sophomore
    Join Date
    Jan 2008
    Posts
    130
    Hey DrRocket,

    Sorry for the late reply, been busy on a job... Sure, thanks for suggesting the book, I will see if I can find it at the library!

    Cheers!
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •