Let \(A=(a_{ij})\) be a differentiable real matrix-valued function with \(\det A>0\), and let \(B=(b_{ij})=A^{-1}\). Prove \[ \frac d{dt}\log\det A=\sum_{i,j=1}^n\frac{da_{ij}}{dt}b_{ji}. \]
Solution
Because \[ \det A(t)>0, \] the matrix \(A(t)\) is invertible and the real logarithm of its determinant is defined.
1. With \(C_{ij}\) the \((i,j)\) cofactor of \(A\), \(\frac{d}{dt}\det A=\sum_{i,j}C_{ij}\frac{da_{ij}}{dt}\).
Let \(C_{ij}\) denote the \((i,j)\) cofactor of \(A\). The determinant is a polynomial in the entries \(a_{ij}\), and \[ \frac{\partial}{\partial a_{ij}}\det A=C_{ij}. \] Thus the chain rule gives \[ \frac d{dt}\det A =\sum_{i,j=1}^n C_{ij}\frac{da_{ij}}{dt}. \]
2. \(C_{ij}=(\det A)\,b_{ji}\).
The adjugate formula says \[ A^{-1}=\frac{1}{\det A}\operatorname{adj}(A), \] and the \((j,i)\) entry of \(\operatorname{adj}(A)\) is \(C_{ij}\). Since \(B=A^{-1}\), \[ b_{ji}=\frac{C_{ij}}{\det A}. \] Equivalently, \[ C_{ij}=(\det A)b_{ji}. \]
3. \(\frac{d}{dt}\log\det A=\sum_{i,j}\frac{da_{ij}}{dt}b_{ji}\).
By steps <1>1–<1>2, \[ \begin{aligned} \frac d{dt}\log\det A &=\frac1{\det A}\frac d{dt}\det A\\ &=\frac1{\det A} \sum_{i,j}C_{ij}\frac{da_{ij}}{dt}\\ &=\sum_{i,j} \frac{da_{ij}}{dt}b_{ji}. \end{aligned} \] Hence \[ \boxed{ \frac d{dt}\log\det A =\sum_{i,j=1}^n\frac{da_{ij}}{dt}b_{ji}.} \]