1. ## 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.

2. ## 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.

3. ## Re: Binary Operation

Originally Posted by abhistud
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$What does it mean? 4. ## Re: Binary Operation It is a little odd that you refer to$\displaystyle A_1$and$\displaystyle A_2$but then have an equation in A and B! 5. ## Re: Binary Operation No, they are not equal. So, the operation is not commutative. 6. ## Re: Binary Operation Originally Posted by Plato Examine with$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.

7. ## Re: Binary Operation

Originally Posted by abhistud
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 ?