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Homework Help: Matrices, how to get this equality

  1. Apr 28, 2008 #1
    How do I get the right side of the equation to equal the left side?


    [tex]

    \begin{bmatrix}
    1 & 0 \\
    0 & 1
    \end{bmatrix} =
    \begin{bmatrix}
    a & 1 \\
    1 & 1
    \end{bmatrix}

    [/tex]

    I tried multiplying the first row by 0 to get rid of a is this is the right first step?
     
  2. jcsd
  3. Apr 28, 2008 #2

    Dick

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    They aren't equal because 0 is not equal to 1. Face it. Massaging it won't change that.
     
  4. Apr 29, 2008 #3
    Sorry I should have explained a bit better.

    What i'm trying to do is find the inverse matrix of
    ax + y = 1
    x + y = b

    I don't think that equal sign should be in there
     
  5. Apr 29, 2008 #4

    Dick

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    Then multiplying the first row by 0 seems like a pretty poor idea. Now you've got NOTHING in the first row. How about multiplying by 1/a, so you can subtract the first row from the second and get a 0 in the lower left part of the original matrix?
     
  6. Apr 29, 2008 #5

    HallsofIvy

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    So your question is "How do I find the inverse to
    [tex]\left[\begin{array}{cc}a & 1 \\ 1 & 1 \end{array}\right][/tex]?

    One way to do that is to use, directly, the definitiion: Find x, y, z u, so that
    [tex]\left[\begin{array}{cc}a & 1 \\ 1 & 1 \end{array}\right][left[\begin{array}{cc}x & y \\ z & u\end{array}\right]= \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right][/tex].

    That is the same as solving the equation ax+ z= 1, ay+ u= 0, x+ z= 0, y+ u= 1 for x, y, z, u.

    But I suspect you are talking about "row reduction". The "row operations" are
    1) Swap two rows
    2) Multiply one row by a number
    3) Add a multiple of one row to another.

    In order to row reduce [itex]\left[\begin{array}{cc}a & 1 \\ 1 & 1 \end{array}\right] do the following:

    Normally, the simplest way to get a "1" on the diagonal is to multiply the entire row by 1 over whatever is on the diagona. Here, since we already have a '1' directly below a, I would swap the first and second rows to get
    [tex]\left[\begin{array}{cc} 1 & 1 \\ a & 1\end{array}\right][/tex]
    To get a "0" below that, subtract a times the first row from the second (do you see how I decided to use "a time"? Since we have a "1" just above the"a", we use a/1= a.)
    [tex]\left[\begin{array}{cc} 1 & 1 \\ 0& 1-a\end{array}\right][/tex]
    Now you want to get a "1" on the diagonal, in the lower right. Since that is 1- a now, the obvious thing to do is divide that row by 1- a. Notice that does not change the first row:
    [tex]\left[\begin{array}{cc} 1 & 1 \\ 0& 1\end{array}\right][/tex]
    Finally, subtract the second row from the first and you are done:
    [tex]\left[\begin{array}{cc} 1 & 0 \\ 0& 1\end{array}\right][/tex]
    Now if you apply those same row operations to the identity matrix, in the same order, you get the inverse matrix. Of course, it is simpler to write the matrices A and the identity matrix side-by-side and do the row operations to both matrices at the same time.
     
  7. Apr 30, 2008 #6
    thanks
     
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