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3x3 matrices

  1. Jul 25, 2008 #1
    So each of three formulas has three variables. I am trying to picture this geometrically as if it were position variables in 3D space. Then each formula describes a unique point in 3D space, and the row vector goes from the 000 point to the point described by the formula. So, there are three points, a triangle, sitting in 3D space, each of the three vertices forming a vector from the point of origin.

    So I begin to use fundamental row operations to get the matrix in row echelon form. What happens to my geometric triangle? Substituting one row for another just changes the axis labels....multiplying one row by a scaler makes the row vector longer. Adding two row vectors and substituting the result back into one of the rows...I am having a bit of difficulty making a picture of that.

    Then, I solve the matrix for the three variables algebraically, and get another end point to make a vector with the origin.

    I am thinking of each vector as a push on an object....the push of the solution vector does the same thing to the object as the three row vectors would, if it replaced them. So, how do I get from flipping triangles to the meaning of the algebraic solution? My first guess is that someone may tell me I am going about it all wrong. I don't like to memorize math manipulations without knowing what they do, geometrically. Surely this should work, at least in 3D?

    Any help appreciated, thanks.
  2. jcsd
  3. Jul 25, 2008 #2
    If you think of the 3 vectors as forming 3 edges of a Parallelepiped, then replacing row 1 with row 1 + row 2 can be seen as shearing the parallelepiped in along the direction of row 2. Same thing if you are interested in the vectors making up the columns.

    This is easiest to see in 2D where it is just a parallelogram. Imagine the matrix is the identity, and you add row 1 to row 2 and put it in row 2. This takes a square and shears it in the x direction creating a parallelogram.

    Now, what does the matrix do to a collection of vectors (Mv) in space? If you work out the multiplication, it turns out that M takes the unit cube, and maps it to the parallelepiped created by the column vectors inside the matrix. The x-axis goes to the first column vector in M, the y axis goes to the second column vector in M, and the z axis goes to the third column vector, and the rest of space warps stretches, rotates, etc accordingly so that it all stays linear.

    Here is a diagram:
    http://img528.imageshack.us/img528/3225/matrixtransformationak0.png [Broken]
    Last edited by a moderator: May 3, 2017
  4. Jul 26, 2008 #3
    Thanks, maze. I think I see that now.

    Then what are we doing when we put the matrix in row echalon form? And what is the geometric meaning of the solution set?
  5. Jul 28, 2008 #4
    Ok, so solving a system of linear equations is the same thing as inverting a matrix. If v=(x,y,z), and M is a 3x3 matrix, then a system of linear equations could be written as Mv=d where d is the vector of solutions. If you multiply things out it is the same. Therefore the solution vector v is given by v=M^-1 d.

    But how do you invert a matrix? The standard method is to do gaussian elimination on M in order to transform it into I, and simultaneously do the same gaussian eliminations on I, transforming it into M^-1.

    Geometrically, why does this work? The gaussian elimination takes your parallelepiped for M, and skews/scales it until it is back to a unit box. However, if you apply the same gaussian elimination to the unit box, it stretches and scales the unit box the exact same, and so you get the shape that *would have* been mapped to the unit box by M. In other words, the inverse.

    http://img150.imageshack.us/img150/1435/rowopinversedv8.png [Broken]
    Last edited by a moderator: May 3, 2017
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