$\det(\exp X)=e^{\mathrm{Tr}\, X}$ for 2 dimensional matrices

Question

I want to prove that for $X\in M_2(\mathbb{R})$ the formula $\det(\exp X)=e^{\mathrm{Tr}\, X}$ holds, writing $X$ in normal form gives $X=PJP^{-1}$, where $J$ is the Jordan matrix, now $\exp (PJP^{-1})=P(\exp J)P^{-1}$ and $\det P(\exp J)P^{-1}=\det \exp J$. If $J=\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ where we allow $a=b$, we get that $\det \exp J=e^{a+b}=e^{\mathrm{Tr}\, J}=e^{\mathrm{Tr}\, X}$, since $X$ and $J$ are similar. However, if $J=\begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}$, then $\exp J=\exp\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \exp \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} e^a & e^{a+1} \\ e^a & e^{a} \end{pmatrix}$, so $\det \exp J=e^{2a}(1-e)$, which is not the result I want.

Answer 1

Your calculation of $\exp J$ is incorrect. $$\exp J = \exp \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \exp \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} e^a & 0 \\ 0 & e^a \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} e^a & e^a\\ 0 & e^a \end{pmatrix}$$

Answer 2

$$ \exp\begin{pmatrix}0&1\\0&0\end{pmatrix}=I+\begin{pmatrix}0&1\\0&0\end{pmatrix}+\frac1{2!}\begin{pmatrix}0&1\\0&0\end{pmatrix}^2+\ldots =I+\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}1&1\\0&1\end{pmatrix}. $$