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  "body": "\n## Proof\n\n1. Let row $i$ of $A$ be zero.\n2. Multiplying row $i$ of $A$ by $0$ gives a matrix whose row $i$ is $0$ and whose other rows are those of $A$. This matrix is $A$ itself.\n3. By the rule for [[scaling a row|009CY8JXFX]] with $c = 0$, we get $\\det A = 0 \\cdot \\det A$.\n4. So $\\det A = 0$.\n\n## Idea\n\nA zero row is unchanged when it is multiplied by $0$. Since the determinant is linear in that row, it must be multiplied by $0$ too.\n",
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