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  "body": "\n## Proof\n\nWe show that $w = \\det \\Phi$ satisfies $w' = (\\tr A)\\, w$ and then solve this equation.\n\n1. Write $\\Phi = (\\varphi_{ij})$ and $A = (A_{ij})$. Each $\\varphi_{ij}$ is differentiable on $I$. By the [[Leibniz formula|016PE4VZ8Q]], $\\det \\Phi = \\sum_{\\sigma} \\operatorname{sgn}(\\sigma) \\prod_{i=1}^{n} \\varphi_{i, \\sigma(i)}$, summed over all permutations $\\sigma$.\n2. By induction on the number of factors, the [[product rule|0V9A43H7WV]] gives $(f_1 \\cdots f_n)' = \\sum_{k} f_1 \\cdots f_k' \\cdots f_n$ for differentiable $f_1, \\dots, f_n$.\n3. With [[linearity of differentiation|0M0TKFRX2A]], $(\\det \\Phi)' = \\sum_{k=1}^{n} \\det \\Phi_k$. Here $\\Phi_k$ is $\\Phi$ with row $k$ replaced by row $k$ of $\\Phi'$.\n4. From $\\Phi' = A \\Phi$, row $k$ of $\\Phi'$ has entries $\\varphi_{kj}' = \\sum_{l} A_{kl} \\varphi_{lj}$. So it is $\\sum_l A_{kl}$ times row $l$ of $\\Phi$.\n5. Each term of the Leibniz formula contains exactly one entry of row $k$. So the determinant is linear in row $k$, and $\\det \\Phi_k = \\sum_l A_{kl} \\det \\Phi_{k, l}$. Here $\\Phi_{k, l}$ is $\\Phi$ with row $k$ replaced by row $l$ of $\\Phi$.\n6. For $l \\ne k$, the matrix $\\Phi_{k, l}$ has two equal rows, so [[its determinant is $0$|0K4VQB7HWR]]. For $l = k$, $\\Phi_{k, k} = \\Phi$. Hence $\\det \\Phi_k = A_{kk} \\det \\Phi$.\n7. Summing over $k$ and using the definition of the [[trace|07PTPSZPME]], $w' = (\\tr A)\\, w$ on $I$, where $w = \\det \\Phi$.\n8. The function $\\tr A$ is continuous, as a sum of continuous entries. Let $T(t) = \\int_{t_0}^{t} \\tr A(s)\\, ds$. By [[the antiderivative theorem|0NC2A3HWGT]], $T' = \\tr A$, and $T(t_0) = 0$.\n9. Let $g = w\\, e^{-T}$. By the product rule, the [[chain rule|0RXX2PT1MQ]] and [[the derivative of the exponential|0EX9GFJ0B3]], $g' = (\\tr A)\\, w\\, e^{-T} - w\\, (\\tr A)\\, e^{-T} = 0$.\n10. So [[$g$ is constant|003QDZ4T16]] on $I$, and $g(t) = g(t_0) = \\det \\Phi(t_0)$.\n11. By the [[exponential of a sum|0E8HNVBCSP]], $e^{T(t)} e^{-T(t)} = 1$. Multiplying $g(t)$ by $e^{T(t)}$ gives $\\det \\Phi(t) = \\det \\Phi(t_0) \\exp\\left(\\int_{t_0}^{t} \\tr A(s)\\, ds\\right)$.\n\n## Idea\n\nDifferentiating a determinant row by row gives a sum of $n$ determinants. In the $k$-th one, the new row $k$ is a combination of the rows of $\\Phi$, and only the multiple $A_{kk}$ of row $k$ survives. So $\\det \\Phi$ solves a scalar linear equation with coefficient $\\tr A$, whose solutions are exponentials.\n",
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