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  "body": "\n## Statement\n\nLet $\\mathbb{K}$ be $\\R$ or $\\C$, let $I \\subseteq \\R$ be an open interval, and let $A \\colon I \\to \\mathbb{K}^{n \\times n}$ be a [[differentiable|0K57YG3BB8]] map.\nThen $t \\mapsto \\det A(t)$ is differentiable, and for all $t \\in I$\n$$\\frac{d}{dt} \\det A(t) = \\tr\\big(\\operatorname{adj}(A(t))\\, A'(t)\\big),$$\nwhere $\\det$ is the [[determinant|00KDZRW4C0]], $\\operatorname{adj}$ is the [[adjugate|07W9H8QF38]] and $\\tr$ is the [[trace|0XC1RDBY7A]].\nIf $A(t)$ is invertible, this equals $\\det A(t) \\cdot \\tr\\big(A(t)^{-1} A'(t)\\big)$.\n\n## Remarks\n\nThe determinant is a polynomial in the entries $a_{ij}$.\nBy the [[Laplace expansion|0C08BQ3M05]] along row $i$, its partial derivative with respect to $a_{ij}$ is the [[cofactor|0XPKR18APA]] $C_{ij}$.\nThe chain rule gives $\\frac{d}{dt}\\det A = \\sum_{i,j} C_{ij}\\, a_{ij}'$.\nThis sum is $\\tr(\\operatorname{adj}(A) A')$, because $\\operatorname{adj}(A)_{ji} = C_{ij}$.\nThe second form follows from $\\operatorname{adj}(A) = \\det(A) A^{-1}$.\n\nA special case is $\\frac{d}{dt}\\det(I + tX)\\big|_{t=0} = \\tr X$: the derivative of the determinant at the identity is the trace.\nFor $A(t) = e^{tX}$, the formula gives $\\frac{d}{dt}\\det e^{tX} = \\tr(X) \\det e^{tX}$, so $\\det e^{tX} = e^{t \\tr X}$.\n\n## Explanations\n\n### L1 human\n\nJacobi's formula tells how fast the determinant of a matrix changes when the matrix changes.\nNear the identity matrix, the rate of change of the determinant is the trace of the change.\nFor example, $\\det\\begin{pmatrix} 1 + t \u0026 0 \\\\ 0 \u0026 1 + 2t \\end{pmatrix} = 1 + 3t + 2t^2$ has derivative $3$ at $t = 0$.\nThis is the trace of $\\begin{pmatrix} 1 \u0026 0 \\\\ 0 \u0026 2 \\end{pmatrix}$.\n\n### L2 human\n\nThe formula says that the differential of $\\det$ at $A$ is the linear map $X \\mapsto \\tr(\\operatorname{adj}(A) X)$.\nIt gives Liouville's formula for the Wronskian of a linear system of differential equations.\nIt also gives the [[determinant of a matrix exponential|0A1QQN333A]], $\\det e^A = e^{\\tr A}$.\nIn Lie theory it says that the derivative of $\\det \\colon \\mathrm{GL}_n(\\mathbb{K}) \\to \\mathbb{K}^\\times$ at the identity is the trace.\n",
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