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Calculate $A \cdot B$ and $B \cdot A$ for $A= \begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -1\\ 1 & 1 \end{pmatrix}$

Is $A \cdot B=B \cdot A$?

This is a task from an old exam and I'd like to know if I did it correctly?

I will only ask for $A \cdot B$ because the other is done the same way.

        |  1     -1
  A*B   |  1      1
---------------------
1    1  | c_11   c_12 
1    1  | c_21   c_22

$c_{11}= 1 \cdot 1+1 \cdot 1=2$

$c_{12}= 1 \cdot (-1)+1 \cdot 1=0$

$c_{22}= 1 \cdot 1+1 \cdot 1=2$

$c_{22}= 1 \cdot (-1) +1 \cdot 1=0$

Thus, $A \cdot B = \begin{pmatrix} 2 & 0\\ 2 & 0 \end{pmatrix}$

Is it correct? I would do $B \cdot A$ this way too and then check if they are equal to complete the task.

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    Yes you are correct. – 2017-01-08
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    Yes, seems correct – 2017-01-08
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    Thank you very much glad to hear! :) – 2017-01-08

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Your calculation is correct.

Notice that if you were not required to calculate $B \cdot A$ then you would not have needed to calculate it entirely in order to get that $A \cdot B \neq B \cdot A$.

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    But why not? :D There is a trick? – 2017-01-08
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    @cnmesr To show they are different it suffices to exhibit that at least one of the four entries is different. So if you knew in advance where to look, you could save work ... Incidentally, computing only the top-left entry of $AB$ and the top-left entry of $BA$ suffices. **But:** The problem statement explicitly asks you to compute $AB$ and $BA$ – 2017-01-08