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Assuming it is about matrices in $\mathbb{R}$ and not taking edge cases into consideration.

Intuitively it is true that for $(\mathbb{F}^{2\times 2}, \cdot)$ commutative property doesn't hold true.

But I'm struggling to prove it with help of definition of Field. Help on it would be greatly appreciated.

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    $\begingroup$ Well, how would you prove it for $\mathbb{R}$? Then you can see whether that proof works for a general field. $\endgroup$ Commented Nov 23, 2022 at 1:39
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    $\begingroup$ every field has $0$ and $1.$ Try pairs of matrices with those entries. $\endgroup$
    – Will Jagy
    Commented Nov 23, 2022 at 1:45
  • $\begingroup$ Do not use x for $\times$. Use \times. $\endgroup$ Commented Nov 23, 2022 at 1:58

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Consider the matrices $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Their commutator $[A, B] = AB - BA = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, which can only be $0$ if $0 = 1$. But that cannot happen in a field.

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