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  "body": "\n## Proof\n\n1. Write the matrix as $A = (a_{ij})$ with $a_{11} = a$, $a_{12} = b$, $a_{21} = c$ and $a_{22} = d$.\n2. By the [[Leibniz formula|016PE4VZ8Q]], $\\det A = \\sum_{\\sigma \\in S_2} \\sgn(\\sigma)\\, a_{1, \\sigma(1)} a_{2, \\sigma(2)}$.\n3. The group $S_2$ has two elements: the identity $\\iota$ and the [[transposition|0FWVRHBRWM]] $\\tau$ that exchanges $1$ and $2$.\n4. By the [[definition of the sign|0SK90238BH]], $\\sgn(\\iota) = (-1)^0 = 1$ and $\\sgn(\\tau) = (-1)^1 = -1$.\n5. The term of $\\iota$ is $a_{11} a_{22} = ad$. The term of $\\tau$ is $-a_{12} a_{21} = -bc$.\n6. So $\\det A = ad - bc$.\n\n## Idea\n\nFor $n = 2$ there are only two permutations. The identity gives the product of the main diagonal, and the swap gives minus the product of the other diagonal.\n",
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