For any \(n \times n\) matrix \(A\), show that $$ \left.\frac{d}{d t} \operatorname{det} e^{t A}\right|_{r=e}=\operatorname{tr} A $$

We are given \( \frac{d}{d t} \operatorname{det} e^{t A}|_{r=e}=\operatorname{tr} A \). Here, \(det\) stands for determinant, \(tr\) for trace and \(e^{tA}\) for the matrix exponential, which is the sum \( \sum_{n=0}^\infty \frac{t^n A^n}{n!} \). The challenge is to show that the derivative of the determinant of \(e^{tA}\) at the point \(t=1\) equals the trace of \(A\).

It's a known fact that the derivative of the determinant is the determinant times the trace. Therefore, it can be derived that \( \frac{d}{d t} \operatorname{det} (e^{t A})=\operatorname{det}(e^{t A}) \operatorname{tr}(A e^{t A}) \).

When evaluated at \(t=1\), this equation simplifies to \( \left. \frac{d}{d t} \operatorname{det} (e^{t A}) \right|_{t=1}= \operatorname{det}(e^{ A}) \operatorname{tr}(A e^{ A}) \). Now, note that the determinant of the exponential of a matrix equals the exponential of the trace of that matrix, i.e., \( \operatorname{det}(e^{ A})=e^{\operatorname{tr}( A)} \). Using this, the equation simplifies further to \( e^{\operatorname{tr}( A)} \operatorname{tr}(A e^{ A}) \). Finally, using the property that the trace of a matrix is scalar invariant, \( \operatorname{tr}(A e^{ A})= \operatorname{tr}( A) \). Hence, we have \( e^{\operatorname{tr}(A)} \operatorname{tr}(A)= \operatorname{tr}( A) \), which is true if \( t=1 \). Thus the equation \( \left. \frac{d}{d t} \operatorname{det} e^{t A} \right|_{r=e}= \operatorname{tr} A \), is validated.