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  "body": "\n## Counterexample\n\nLet $A = \\begin{pmatrix} 0 \u0026 1 \\\\ 0 \u0026 0 \\end{pmatrix}$ and $B = \\begin{pmatrix} 0 \u0026 0 \\\\ 1 \u0026 0 \\end{pmatrix}$, as real $2 \\times 2$ matrices.\nThen\n$$AB = \\begin{pmatrix} 1 \u0026 0 \\\\ 0 \u0026 0 \\end{pmatrix}, \\qquad BA = \\begin{pmatrix} 0 \u0026 0 \\\\ 0 \u0026 1 \\end{pmatrix}.$$\nThus $AB \\ne BA$.\nThis shows that [[matrix multiplication|032Q9FB8MT]] of $n \\times n$ matrices is not commutative for $n = 2$.\n\n## Remarks\n\nThe same example, extended with zero rows and zero columns, shows that $M_n(R)$ is not commutative for each $n \\ge 2$ and each nonzero ring $R$.\nTwo matrices $A$ and $B$ with $AB = BA$ are said to *commute*.\nSome pairs commute, for example a matrix and the [[identity matrix|06DW9BAWB1]], or two [[diagonal matrices|044JNC7ZAA]].\n\n## Explanations\n\n### L1 human\n\nFor numbers, $2 \\cdot 3 = 3 \\cdot 2$, but for matrices the order of the factors matters.\nHere $AB$ and $BA$ are different matrices.\nA reason is that $AB$ means \"first apply $B$, then $A$\", and two actions done in a different order can give different results.\n",
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