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Thread: Binary Operation

  1. #1
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    Binary Operation

    Let ∗ be a binary operation on the set M2(R) of all 2 2 matrices defined by∀A1, A2 ∈ M2(R), A ∗ B =1/2(AB-BA)
    (a) Prove that the operation is binary.(b) Determine whether the operation is associative and/or commutative.

    I think its a binary operation because R X R = R.
    But, I am unable to prove the second part.
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  2. #2
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    Re: Binary Operation

    I don't know what A1, A2 are but it sounds like you have the set of all 2x2 matrices with real elements. Is that correct?

    If so for part (a) you need to prove the operation A*B produces a 2x2 matrix with real elements.

    to do part (b) you need to show

    i) $A*B = B*A \Rightarrow \dfrac 1 2 (AB-BA) = \dfrac 1 2 (BA - AB)$ ; and

    ii) $(A*B)*C = A*(B*C),~\forall A,B,C \in M_2(\mathbb{R})$

    all of this is pretty straightforward matrix arithmetic.
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  3. #3
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    Re: Binary Operation

    Quote Originally Posted by abhistud View Post
    Let ∗ be a binary operation on the set M2(R) of all 2 2 matrices defined by∀A1, A2 ∈ M2(R), A ∗ B =1/2(AB-BA)
    (a) Prove that the operation is binary.(b) Determine whether the operation is associative and/or commutative.
    I think its a binary operation because R X R = R.
    But, I am unable to prove the second part.
    Examine with $A = \left( {\begin{array}{*{20}{c}} 1&{ - 1} \\ { - 1}&1 \end{array}} \right)~\&~B = \left( {\begin{array}{*{20}{c}} { - 1}&{ - 1} \\ 1&1 \end{array}} \right),$ $A*B~\&~B*A.$
    Are they equal$ What does it mean?
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  4. #4
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    Re: Binary Operation

    It is a little odd that you refer to A_1 and A_2 but then have an equation in A and B!
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    Re: Binary Operation

    No, they are not equal. So, the operation is not commutative.
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    Re: Binary Operation

    Quote Originally Posted by Plato View Post
    Examine with $A = \left( {\begin{array}{*{20}{c}} 1&{ - 1} \\ { - 1}&1 \end{array}} \right)~\&~B = \left( {\begin{array}{*{20}{c}} { - 1}&{ - 1} \\ 1&1 \end{array}} \right),$ $A*B~\&~B*A.$
    Are they equal$ What does it mean?
    No, they are not equal. So, the operation is not commutative.
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  7. #7
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    Re: Binary Operation

    Quote Originally Posted by abhistud View Post
    No, they are not equal. So, the operation is not commutative.
    CORRECT!
    $ \begin{align*}(A*B)* C&=(A*B)C-C(A*B)\\&=(AB-BA)C-C(AB-BA)\\&=ABC-BAC-CAB+CBA\end{align*}$

    Now $ \begin{align*}A*(B* C)&=A(B*C)-(B*C)A\\&=A(BC-CB)-(BC-CB)A\\&=ABC-ACB-BCA+CBA\end{align*}$

    Can you find an example for $C$ where those are not equal ?
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