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Find all 2x2 matrices such that A=A^-1

  1. Jul 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Find all 2x2 matrices such that [tex]A=A^-^1[/tex] (the inverse, just in case the notation is different)

    2. Relevant equations

    [tex]

    A=
    \begin{bmatrix}
    a & b
    \\ c & d
    \end{bmatrix}
    [/tex]

    3. The attempt at a solution

    This is my second attempt at this question. The first time, I took a different approach, using the four equations below. This time, I started off with four other different equations and had a very tough time trying to figure it all out. If I hadn't looked at the solutions, I never would've gotten anywhere. Even with it, I struggled a lot and it took an unreasonable amount of time to work it all out...

    ----

    If [tex]A=A^-1[/tex], then [tex]A^2=I[/tex]. From that, I derived the following equations:

    (i)[tex]a^2+bc=1[/tex]
    (ii)[tex]ab+bd=0[/tex]
    (iii)[tex]ac+cd=0[/tex]
    (iv)[tex]d^2+bc=1[/tex]

    From (ii) and (iii), if a≠-d, then b=0 and c=0. Then from (i) and (iv),a^2=1 and d^2=1and since a≠-d, then a=1 and d=1 or a=-1 and d=-1

    So from that, I get two 2x2 matrices

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

    \begin{bmatrix}
    -1 & 0
    \\ 0&-1
    \end{bmatrix}
    [/tex]


    Then to get the others, I let b=0 and c be any real number (again, I never would've done this if I hadn't seen the answers), then to satisfy (iii), a and d would have to be opposite. I did that for c=0 as well and the result was four more matrices

    [tex]
    \begin{bmatrix}
    1 & 0
    \\ k &-1
    \end{bmatrix}

    \begin{bmatrix}
    -1 & 0
    \\ k &1
    \end{bmatrix}

    \begin{bmatrix}
    1 & k
    \\ 0 & -1
    \end{bmatrix}

    \begin{bmatrix}
    -1 & k
    \\ 0&1
    \end{bmatrix}
    [/tex]

    To get the last answer given in the book:

    [tex]

    \begin{bmatrix}
    a & b
    \\ \frac{1-a^2}b&-a
    \end{bmatrix}
    [/tex]

    I just rearranged (i) to get b. But couldn't you write represent c in a similar way but with b in the denominator? Are they just writing it in terms of a and b?

    Anyway, though it's solved, it took me super long and I'm still not happy with the method I used. I went down another road where I said b=0 iff a+d≠0 because I used

    [tex]b=\frac{-b}{ad-bc}[/tex]

    [tex]ab+bd=0[/tex]
    [tex]b(a+d)=0[/tex]
    [tex]b=0[/tex]

    And I also said that if b≠0, then ad-bc-1 and I had all these equations but couldn't get anywhere with them.

    Basically, I just want to see how you guys would approach this, what would be a systematic way to solve it? This wasn't even a challenge question and I got destroyed by it.

    EDIT: Also, I'm unsure about my reasoning. I was more happy with the way I did it last time.
     
  2. jcsd
  3. Jul 5, 2012 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Subtract equation (iv) from equation (i).

    That gives [itex]d=\pm\,a\ .[/itex]

    What do equations (ii) & (iii) tell you ?
     
  4. Jul 5, 2012 #3
    Well there is a quick way of finding the inverse of a 2x2 matrix

    [tex]A=
    \begin{bmatrix}
    a & b
    \\ c & d
    \end{bmatrix}[/tex]

    [tex]A^{-1}=\frac{1}{det(A)}*adj(A)[/tex]

    and the adjugate of a 2x2 matrix is just

    [tex]
    \begin{bmatrix}
    d & -b
    \\ -c & a
    \end{bmatrix}[/tex]

    the determinant is [tex]ad-bc[/tex]

    which by equating elements of the matrix and it's inverse you can show [tex]ad-bc=-1[/tex] and that [tex]d=-a[/tex]

    and from here getting the rest of the solution isn't too complicated.
     
  5. Jul 6, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Something you will find useful over and over again is a simple rule for inverting a 2x2 matrix:
    swap the diagonal elements, change sign of the off-diagonal elements, and divide by the determinant. (Others have displayed this for you, but expressing it in words may help.)

    RGV
     
  6. Jul 6, 2012 #5

    I like Serena

    User Avatar
    Homework Helper

    Hi autodidude! :smile:

    If you're interested in alternative and simpler ways to solve it, here's another (although it does require some advanced knowledge :wink:).


    Since you have ##A^2-I=0##, according to Cayley–Hamilton, the eigenvalues of A must be roots of ##x^2-1=0##.
    So the eigenvalues can only be ##\pm 1##.


    And according to Jordan, the possible matrices A must then be "similar" to one of:
    $$
    J = \begin{bmatrix}1&0\\0&1\end{bmatrix},\

    \begin{bmatrix}-1&0\\0&-1\end{bmatrix},\

    \begin{bmatrix}1&0\\0&-1\end{bmatrix},\

    \begin{bmatrix}1&1\\0&1\end{bmatrix},\

    \begin{bmatrix}-1&1\\0&-1\end{bmatrix}
    $$
    However not all of these matrices are a solution: the last 2 do not "fit", which you can verify by calculating ##A^2##, or by finding the inverse as mentioned before (swapping the entries on the main diagonal, negating the other 2 entries, and dividing by the determinant).


    The term "similar" means that for any invertible matrix B, the matrix product ##A = B J B^{-1}## is a solution.

    In the cases that J=I or J=-I, you can see that the corresponding matrices simplify to ##A=B I B^{-1}=I##, respectively ##A=B \cdot -I \cdot B^{-1}=-I##.


    So the solutions with any invertible matrix B are:
    $$
    A = I,\

    A = -I,\

    A = B \begin{bmatrix}1&0\\0&-1\end{bmatrix} B^{-1}
    $$
     
    Last edited: Jul 6, 2012
  7. Jul 12, 2012 #6
    I did manage to get ad-bc = -1 the second time I did it. I think the part that wasn't supposed to be complicated turned out to be very complicated for me :)

    @I like Serena: I'm very interested in other methods...but yeah, that's beyond where I'm at the moment. Looks much nicer :p
     
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