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Math Help - Group problem with matrices

  1. #1
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    Group problem with matrices

    Let G = \left[\begin{array}{cc}a&a\\a&a\end{array}\right] | a \in R, a \ne 0. Show that G is a group under matrix multiplication. Explain why each element of G has an inverse even though the matrices have 0 determinant.

    Why isn't my latex code working? I want to produce a matrix with two rows, each rows with [ a a ]...
    Last edited by mr fantastic; July 10th 2011 at 04:32 PM.
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  2. #2
    MHF Contributor red_dog's Avatar
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    \displaystyle G=\left\{X_a=\left.\left(\begin{array}{cc}a & a\\a & a\end{array}\right)\right|a\in\mathbf{R},a\neq 0\right\}

    Let X_a,X_b\in G\Rightarrow X_a\cdot X_b=\left(\begin{array}{cc}2ab & 2ab\\2ab & 2ab\end{array}\right)=X_{2ab}\in G
    We have X_a=X_b\Leftrightarrow a=b
    Comutativity:
    X_a\cdot X_b=X_{2ab}=X_{2ba}=X_b\cdot X_a
    Asociativity:
    (X_a\cdot X_b)\cdot X_c=X_{2ab}\cdot X_c=X_{4abc}
    X_a\cdot(X_b\cdot X_c)=X_a\cdot X_{2bc}=x_{4abc}
    Then (X_a\cdot X_b)\cdot X_x=X_a\cdot(X_b\cdot X_c)
    Identity element:
    Let X_e be the identity element:
    Then X_a\cdot X_e=X_e\cdot X_a=X_a
    We have X_a\cdot X_e=X_a\Leftrightarrow X_{2ae}=X_a\Leftrightarrow 2ae=a\Leftrightarrow e=\frac{1}{2}
    So the identity element is \displaystyle X_{\frac{1}{2}}=\left(\begin{array}{cc}\frac{1}{2} & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2}\end{array}\right)
    Reciprocal element:
    Let X_a\in G and X_{a'}\in G the reciprocal element.
    Then X_a\cdot X_{a'}=X_{a'}\cdot X_a=X_{\frac{1}{2}}
    X_{2aa'}=X_{\frac{1}{2}}\Rightarrow 2aa'=\frac{1}{2}\Rightarrow a'=\frac{1}{4a}
    So, the reciprocal element of X_a is X_{\frac{1}{4a}}

    Each element has an inverse because the identity element is not I_2 and we can't apply the same rule for an inverse of a matrix.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by tttcomrader View Post
    Let G = \left[\begin{array}{cc}a&a\\a&a\end{array}\right] | a \in R, a \ne 0. Show that G is a group under matrix multiplication. Explain why each element of G has an inverse even though the matrices have 0 determinant.

    Why isn't my latex code working? I want to produce a matrix with two rows, each rows with [ a a ]...
    The LaTex problem was not with the array definition but due to not leaving a
    space after \in.

    RonL
    Last edited by mr fantastic; July 10th 2011 at 04:32 PM.
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  4. #4
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    Thanks, that was really helpful, I just need to think a bit more about why each element has an inverse because of the different identity.
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