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Math Help - List Two Different Matrices A s.t. A^2=I

  1. #1
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    List Two Different Matrices A s.t. A^2=I

    Hello,

    I am trying to determine two different matrices A such that A^2=\Bigg[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\Bigg].

    I am trying to solve this algebraically. Here is what I have attempted:



    Last edited by Fourier; September 4th 2007 at 04:47 PM.
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  2. #2
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    Quote Originally Posted by Fourier View Post
    Hello,

    I am trying to determine two different matrices A such that A^2=\Bigg[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\Bigg].

    I am trying to solve this algebraically. Here is what I have attempted:



    Set b = c = 0. Then a^2 = d^2 = 1. So I and -I fall out as 2 of 4 solutions.
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  3. #3
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    Quote Originally Posted by Fourier View Post
    Hello,

    I am trying to determine two different matrices A such that A^2=\Bigg[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\Bigg].

    I am trying to solve this algebraically. Here is what I have attempted:



    Set a^2=d^2=0, then bc=1 so:

    <br />
A^2= \left[ \begin{array}{cc}0&x\\1/x&0 \end{array} \right]^2=I_{2x2} <br />

    RonL
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  4. #4
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    Hello, Fourier!

    You're off to a good start . . .


    . . \begin{bmatrix}a & b\\c &d\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix  }\;=\;\begin{bmatrix}1&0\\0&1\end{bmatrix}

    \begin{array}{ccc}(1)\;\;a^2 + bc \:=\:1 &\qquad & (2)\;\;ab+bd \:=\:0 \\<br /> <br />
(3)\;\;ac + cd \:=\:0 & \qquad & (4)\;\;bc+d^2\:=\:0\end{array}

    We have: . \begin{array}{cccc}(2)\;\;ab+bd\:=\:0 & \Rightarrow & b(a+d)\:=\:0 & (5) \\ (3)\;\;ac + cd \:=\:0 & \Rightarrow & c(a+d)\:=\:0\ & (6)\end{array}


    Subtract: . \begin{array}{c}(1)\;\;a^2+bc \:=\:1 \\ (4)\;\;bc+d^2\:=\:1\end{array} \quad\Rightarrow\quad a^2-d^2\:=\:0\quad\Rightarrow\quad d \,=\,\pm a


    If a\!\cdot\!d \neq 0, then (5) and (6) give us: . b = c = 0

    And (1) and (4) give us: . a^2 \,= \,1,\;d^2\,=\,1\quad\Rightarrow\quad a\,=\,\pm1,\;d\,=\,\pm1

    . . Two solutions: . \begin{bmatrix}1 &0 \\0&1\end{bmatrix} .and . \begin{bmatrix}\text{-}1 & 0\\0&\text{-}1\end{bmatrix}



    If a = d = 0, then (1) gives us: . bc \,=\,1\quad\Rightarrow\quad c \,=\,\frac{1}{b}

    . . More solutions: . \begin{bmatrix}0 & b \\ \frac{1}{b} & 0\end{bmatrix} . . . . for b \neq 0 . obviously.

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, Fourier!

    You're off to a good start . . .



    We have: . \begin{array}{cccc}(2)\;\;ab+bd\:=\:0 & \Rightarrow & b(a+d)\:=\:0 & (5) \\ (3)\;\;ac + cd \:=\:0 & \Rightarrow & c(a+d)\:=\:0\ & (6)\end{array}


    Subtract: . \begin{array}{c}(1)\;\;a^2+bc \:=\:1 \\ (4)\;\;bc+d^2\:=\:1\end{array} \quad\Rightarrow\quad a^2-d^2\:=\:0\quad\Rightarrow\quad d \,=\,\pm a


    If a\!\cdot\!d \neq 0, then (5) and (6) give us: . b = c = 0

    And (1) and (4) give us: . a^2 \,= \,1,\;d^2\,=\,1\quad\Rightarrow\quad a\,=\,\pm1,\;d\,=\,\pm1

    . . Two solutions: . \begin{bmatrix}1 &0 \\0&1\end{bmatrix} .and . \begin{bmatrix}\text{-}1 & 0\\0&\text{-}1\end{bmatrix}



    If a = d = 0, then (1) gives us: . bc \,=\,1\quad\Rightarrow\quad c \,=\,\frac{1}{b}

    . . More solutions: . \begin{bmatrix}0 & b \\ \frac{1}{b} & 0\end{bmatrix} . . . . for b \neq 0 . obviously.

    More solutions are

    \begin{bmatrix}\text{-1} &0 \\0&1\end{bmatrix} and \begin{bmatrix}1 & 0\\0&\text{-}1\end{bmatrix}

    and

    \begin{bmatrix}a & b \\ \frac{1-a^2}{b} & \text{-}a\end{bmatrix} and \begin{bmatrix}a & \frac{1-a^2}{b} \\ b & \text{-}a\end{bmatrix} for b \neq 0.
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