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Math Help - List Matrices

  1. #1
    Senior Member DivideBy0's Avatar
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    List Matrices

    Find all 2 x 2 matrices such that A^{-1}=A.

    Is there a systematic way to do this?
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let \displaystyle A=\left(\begin{array}{ll}a & b\\c & d\end{array}\right),a,b,c,d\in\mathbf{R}
    A=A^{-1}\Leftrightarrow A^2=I_2.
    Then \displaystyle A=\left(\begin{array}{ll}a^2+bc & b(a+d)\\c(a+d) & d^2+bc\end{array}\right)=\left(\begin{array}{ll}1 & 0\\0 & 1\end{array}\right).
    So \displaystyle \left\{\begin{array}{l}a^2+bc=1\\b(a+d)=0\\c(a+d)=  0\\d^2+bc=1\end{array}\right..
    I. If a+d\neq 0\Rightarrow b=c=0\Rightarrow a^2=d^2=1\Rightarrow a=1,d=1 or a=-1,d=-1\Rightarrow
    \displaystyle A=\left(\begin{array}{ll}1 & 0\\0 & 1\end{array}\right)=I_2 or A=\left(\begin{array}{ll}-1 & 0\\0 & -1\end{array}\right)=-I_2.
    II. If a+d=0\Rightarrow d=-a\Rightarrow A=\left(\begin{array}{ll}a & b\\c & -a\end{array}\right), so that a^2+bc=1
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  3. #3
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    This is a rather more straightforward way.
    If  A = \left[ {\begin{array}{lr}<br />
   a & b  \\<br />
   c & d  \\<br />
\end{array}} \right]\quad then A^{ - 1}  = \left[ {\begin{array}{lr}<br />
   {\frac{d}{\Delta }} & {\frac{{ - b}}{\Delta }}  \\<br />
   {\frac{{ - c}}{\Delta }} & {\frac{a}{\Delta }}  \\<br />
\end{array}} \right],\quad \Delta  = ad - bc \not= 0

    That means that if A = A^{ - 1}  \Rightarrow \quad \begin{array}{lr}<br />
   {\frac{d}{\Delta } = a} & {\frac{{ - b}}{\Delta } = b}  \\<br />
   {\frac{{ - c}}{\Delta } = c} & {\frac{a}{\Delta } = d}  \\<br />
\end{array}.

    Thus \Delta  =  - 1\quad  \Rightarrow \quad a =  - d; that any two-by-two matrix with determinant equal –1 and a=-d has the property.
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  4. #4
    Senior Member DivideBy0's Avatar
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    But doesn't a=-d exclude the possibility of the

    [1 0]
    [0 1]

    matrix? or the

    [-1 0]
    [0 -1]

    or are they just exceptions?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    But doesn't a=-d exclude the possibility of the

    [1 0]
    [0 1]

    matrix? or the

    [-1 0]
    [0 -1]

    or are they just exceptions?
    [quote=red_dog;58719]
    I. If a+d\neq 0\Rightarrow b=c=0\Rightarrow a^2=d^2=1\Rightarrow a=1,d=1 or a=-1,d=-1\Rightarrow
    This is one of the two cases that red_dog gave and covers your answer.

    -Dan
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