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  "body": "\n## Proof\n\n1. Write the matrix as $A = (a_{ij})$. Its first row is $a, b, c$, its second row is $d, e, f$ and its third row is $g, h, i$.\n2. By the [[Leibniz formula|016PE4VZ8Q]], $\\det A = \\sum_{\\sigma \\in S_3} \\sgn(\\sigma)\\, a_{1, \\sigma(1)} a_{2, \\sigma(2)} a_{3, \\sigma(3)}$.\n3. The group $S_3$ has six elements. We list each $\\sigma$ as $(\\sigma(1), \\sigma(2), \\sigma(3))$.\n4. The identity $(1, 2, 3)$ is a product of zero [[transpositions|0FWVRHBRWM]]. By the [[definition of the sign|0SK90238BH]], its sign is $1$, and its term is $aei$.\n5. The permutation $(2, 3, 1)$ equals $(1\\,3)(1\\,2)$, a product of two transpositions. Its sign is $1$, and its term is $a_{12} a_{23} a_{31} = bfg$.\n6. The permutation $(3, 1, 2)$ equals $(1\\,2)(1\\,3)$, a product of two transpositions. Its sign is $1$, and its term is $a_{13} a_{21} a_{32} = cdh$.\n7. The permutation $(3, 2, 1)$ is the transposition $(1\\,3)$. Its sign is $-1$, and its term is $-a_{13} a_{22} a_{31} = -ceg$.\n8. The permutation $(2, 1, 3)$ is the transposition $(1\\,2)$. Its sign is $-1$, and its term is $-a_{12} a_{21} a_{33} = -bdi$.\n9. The permutation $(1, 3, 2)$ is the transposition $(2\\,3)$. Its sign is $-1$, and its term is $-a_{11} a_{23} a_{32} = -afh$.\n10. Adding the six terms gives $\\det A = aei + bfg + cdh - ceg - bdi - afh$.\n\n## Idea\n\nFor $n = 3$ the Leibniz formula has $3! = 6$ terms. The identity and the two cyclic shifts are even and give the three products along the down-right diagonals. The three transpositions are odd and give the other three products with a minus sign.\n",
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