Calculate det $(e^A)$ when A= $ \begin{pmatrix} 1 & 0 & 3\\ -1 & 2 & 0 \\ 0 & 1 & -1 \\ \end {pmatrix} $ without calculating det $e^A$.
The matrix A is diagonalizable matrix than i can use the formula det $(e^A)$=$e^{tr(A)}$, but the problem comes when i try to calculate eigenvalues. The characteristic polynomial is: $-λ^3 + 2 λ^2 + λ - 5$ which impossible to solve without wolfarm alpha.
Using Jacobi's formula, one can get the identity: $$ \det(e^{tA}) = e^{\text{tr}(tA)} $$ which is formula you want when $t=1$. This case always holds for any fixed $A$, so you don't need to worry about the characteristic polynomial. This is because you only need $e^A$ to be invertible, which it always is.
Maybe it's also worth noting that, in the real case, you can't generally write any matrix $A$ as $A=e^X$ for some $X$, unless $ \det(A)> 0 $.
Thus, we get: $$ \det(e^A) = e^{\text{tr}(A)} = e^{1+2-1} = e^2$$