Proof of Liouville's Formula
We omit the argument x for brevity. By the Leibniz formula for determinants, the derivative of the determinant of Φ = (Φ_{i, j })_{i, j ∈ {0,...,n}} can be calculated by differentiating one row at a time and taking the sum, i.e.

(2)
Since the matrixvalued solution Φ satisfies the equation, we have for every entry of the matrix Φ'
or for the entire row
When we subtract from the i th row the linear combination
of all the other rows, then the value of the determinant remains unchanged, hence
for every i ∈ {1, . . ., n} by the linearity of the determinant with respect to every row. Hence

(3)
by (2) and the definition of the trace. It remains to show that this representation of the derivative implies Liouville's formula.
Fix x_{0} ∈ I. Since the trace of A is assumed to be continuous function on I, it is bounded on every closed and bounded subinterval of I and therefore integrable, hence
is a well defined function. Differentiating both sides, using the product rule, the chain rule, the derivative of the exponential function and the fundamental theorem of calculus, we obtain
due to the derivative in (3). Therefore, g has to be constant on I, because otherwise we would obtain a contradiction to the mean value theorem (applied separately to the real and imaginary part in the complexvalued case). Since g(x_{0}) = det Φ(x_{0}), Liouville's formula follows by solving the definition of g for det Φ(x).
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