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Using Matrix Multiplication to Obtain Another Matrix

  1. Mar 1, 2013 #1
    1. The problem statement, all variables and given/known data

    I need to use matrix multiplication of matrices A-D to obtain matrix E. I also need to set a equal to some value that would allow me to perform this multiplication.


    2. Relevant equations

    The matrixes I need to multiply:
    A = [1, 1; 0, 1]
    B = [1, 0; 1, 1]
    C = [a, 0; 0, 1]
    D = [1, 0; 0, a]

    By multiplying the above matrices in some order, I need to obtain the following matrix:

    E = [0, 1; 1, 0]

    3. The attempt at a solution

    I have tried setting a = 0, 1, -1, 2, and -2 (since those seemed like the most logical choices) and tried all combinations of multiplication. For the life of me, I can't figure out the order the multiplication should go in and which of the four matrices to multiply together to get matrix E. There has to be a simpler way of approaching the problem, right? I just need some help getting onto the right path.
     
  2. jcsd
  3. Mar 1, 2013 #2

    jbunniii

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    Note that ##CD = DC = aI## (i.e., the scalar ##a## times the identity matrix), and ##AB(aI) = A(aI)B = (aI)AB##. None of these will work, because ##AB## is not a multiple of ##E##. Similarly, ##BA## is not a multiple of ##E##. So you know ##C## and ##D## can't be adjacent.

    Maybe you can make some other observations which will help you reduce the possibilities further. Anything has to be better than just randomly trying different arrangements, because there are a lot of them (##4! = 24##).

    Another observation is that ##E^2 = I##, not sure if that will be helpful.
     
    Last edited: Mar 1, 2013
  4. Mar 1, 2013 #3
    Thanks for your help, but I'm still struggling tremendously. I understand that a will most likely be 0 or -1 (or maybe 1, but that would make C and D the same identity matrix). Is there any other guidance I could get?
     
  5. Mar 1, 2013 #4

    Ray Vickson

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    It might be easier to look at the transformations the matrices induce. If
    [tex] x= \pmatrix{x_1\\x_2}[/tex] then
    [tex] A: x \longrightarrow \pmatrix{x_1+x_2\\x_2}\\
    B: x \longrightarrow \pmatrix{x_1\\x_1+x_2}\\
    C: x \longrightarrow \pmatrix{ax_1\\x_2}\\
    D: x \longrightarrow \pmatrix{x_1\\ax_2}
    [/tex] and you want to find a sequences of such transformations (and a value of 'a') that gives
    [tex] \pmatrix{x_1\\x_2} \: \longrightarrow \pmatrix{x_2\\x_1}[/tex]
     
    Last edited: Mar 1, 2013
  6. Mar 1, 2013 #5

    jbunniii

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    You can use determinants to find the possible value(s) for ##a##.
     
  7. Mar 1, 2013 #6
    Sorry, I still haven't been able to make any progress at all. Thanks anyway.
     
  8. Mar 1, 2013 #7

    jbunniii

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    Not even in finding the possible value(s) for ##a##? Did you try taking determinants of both sides as I suggested? The answer might surprise you: ##a## can't even be a real number, assuming I did the calculation correctly.
     
  9. Mar 1, 2013 #8
    I haven't been taught how to use determinants like you said, and it wouldn't be possible for a to be an imaginary number for this solution. If I wasn't clear, I have to figure out some kind of combination of multiplying matrixes A-D in order to get E as the result. First, I have to figure out what a should be. So, for example, a solution would be (A*B)*(C*D), or some similar combination.

    I've tried everything and I've been working on this for hours. I just want to cry.
     
  10. Mar 1, 2013 #9

    jbunniii

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    I don't think a solution exists. Here is how you use determinants to find ##a##. Regardless of the order in which you multiply ##A##, ##B##, ##C##, and ##D##, the resulting product will have the same determinant, namely ##\det(A)\det(B)\det(C)\det(D)##, where for a general matrix ##[a\textrm{ }b; c\textrm{ }d]##, the determinant is ##ad - bc##. Using this formula, we get ##\det(A) = \det(B) = 1##, ##\det(C) = \det(D) = a##, and ##\det(E) = -1##. As ##\det(A)\det(B)\det(C)\det(D) = \det(E)##, we have ##a^2 = -1##, so ##a = i## or ##a = -i##.

    Out of curiosity, I wrote a brute-force Matlab script which computes the products of all 24 possible permutations, assuming either ##a = i## or ##a = -i##. No permutation even resulted in a real-valued matrix, let alone ##E##. So either I made a coding error or there is no solution.
     
  11. Mar 2, 2013 #10

    Ray Vickson

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    You can do it, but you need more than 4 multiplications. Recall how the matrices transform vectors:
    [tex] A:\pmatrix{x\\y}\rightarrow\pmatrix{x+y\\y}\\
    B: \pmatrix{x\\y} \rightarrow \pmatrix{x\\x+y}\\
    C: \pmatrix{x\\y} \rightarrow \pmatrix{ax\\y}\\
    D: \pmatrix{x\\y} \rightarrow \pmatrix{x\\ay}
    [/tex]
    If we set ##a = -1## we can transform
    [tex]\pmatrix{x\\y} \rightarrow \pmatrix{y\\x} [/tex]
    through a sequence of such moves:
    [tex] \pmatrix{x\\y} \rightarrow \pmatrix{-x\\y} \rightarrow \pmatrix{-x\\y-x}
    \rightarrow \pmatrix{x\\y-x}
    \rightarrow \pmatrix{y\\y-x} \rightarrow \pmatrix{-y\\y-x} \rightarrow \pmatrix{-y\\-x}
    \rightarrow \pmatrix{y\\-x} \rightarrow \pmatrix{y\\x}.[/tex]
     
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