1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Quick basic lin alg quest

  1. Jun 14, 2010 #1
    The problem statement, all variables and given/known data

    Which three matrices E21, E31, E32, put A into triangular form U?

    A =
    [1 1 0]
    [4 6 1]
    [-2 2 0]

    and E32E31E21A = U.

    Multiply those E's to get one matrix M that does elimination: MA = U.

    Attempt at a solution

    I already found all the E's. What I don't get is, if it didn't tell me E32E31E21A = U, how would I know the correct order of multiplication in for M? In other words, how do I know M = E32E31E21 instead of E21E31E32

    edit- Thanks Tedjn, HallsofIvy
     
    Last edited: Jun 14, 2010
  2. jcsd
  3. Jun 14, 2010 #2
    The job of matrix Emn is to place a zero in the m,n position of the matrix. Following the usual row reduction algorithm, it is important that we do the entire first column first, then the second column, etc. (Why is this order important?) The closest matrix to A is E21 because our first step is to place a zero in the 2,1 position, and we perform that multiplication first.
     
  4. Jun 14, 2010 #3

    HallsofIvy

    User Avatar
    Science Advisor

    The simplest way to put a matrix into triangular form is to "row reduce" it. And each "row operation" corresponds to an elementary matrix- the matrix you get by applying that row operation to the identity matrix. Applying a row operation to a matrix is the same as multiplying the matrix by the corresponding elementary matrix- and a series of elementary matrices is multiplied in the same order the row operations were applied.

    For example, here, the first row operation I would use to reduce A to diagonal form would be to subtract four times the first row from the second: that corresponds to elementary matrix
    [tex]\begin{bmatrix}1 & 0 & 0 \\ -4 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]
    and is done by the matrix multiplication
    [tex]\begin{bmatrix}1 & 0 & 0 \\ -4 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 0 \\ 4 & 6 & 1 \\- 2 & 2 & 0\end{bmatrix}[/tex]

    The second thing I would do would be to add two times the first row to the third row:
    [tex]\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ -4 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 0 \\ 4 & 6 & 1 \\- 2 & 2 & 0\end{bmatrix}[/tex]

    It will take a total of 6 row reductions (two per column) to reduce a 3 by 3 matrix to a diagonal matrix. Since you are asked for three matrices, I suspect they intend for you to multiply the two matrices corresponding to a single column together. That is, the first matrix on the right for your matrix multiplication would be
    [tex]\begin{bmatrix}1 & 0 & 0 \\ -4 & 1 & 0 \\ 2 & 0 & 1\end{bmatrix}[/tex]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook