When does $\det e^A=e^{\det A}?$ 
Which $2\times 2$ matrices satisfy the equation
  $$\det e^A=e^{\det A}?$$

I know that $\det e^A=e^{\operatorname{trace}A}$ so assuming $A$ is real we get
$$\operatorname{trace}A=\det A.$$
Then,
$$\det(A-\lambda I)=\lambda^2-2\operatorname{trace}(A)\lambda+\det(A)=(\lambda-\det(A))^2$$
so the only eigenvalue is
$$\lambda=\det(A)=\operatorname{trace}A.$$
Hence,
$$\operatorname{trace}A=2\lambda=2\operatorname{trace}A\implies\operatorname{trace}A=\det A=0.$$
Write
$$A=\begin{pmatrix}a&b\\c&-a\end{pmatrix}$$
so that $\operatorname{trace}A=0$ is already taken into account. Then,
$$\det A=-a^2-bc=0$$
so $a^2=-bc$. Thus,
$$A=\begin{pmatrix}\sqrt{-bc}&b\\c&-\sqrt{-bc}\end{pmatrix}$$
Is that the end of the solution?
 A: The condition $\lambda_1 + \lambda_2 = \lambda_1 \lambda_2$ can be written as
$(\lambda_1 - 1)(\lambda_2 - 1) = 1$.  Thus the pair of eigenvalues $(\lambda_1, \lambda_2)$, if real, are on a hyperbola in the $\lambda_1,\lambda_2$ plane.  You also have solutions where $(\lambda_1, \lambda_2)$ are a pair of complex conjugates $\alpha \pm i \beta$, where
$(\alpha - 1)^2 + \beta^2 = 1$ (this describes a circle in the $\alpha, \beta$ plane).  An example of a real matrix with the real eigenvalues $\lambda_1, \lambda_2$ is
$$ \pmatrix{\lambda_1 & 0\cr 0 & \lambda_2\cr}$$ 
An example of a  real matrix with the complex eigenvalues $\alpha \pm i \beta$ is
$$ \pmatrix{0 & -\alpha^2 - \beta^2\cr 1 & 2\alpha}$$
Of course you way take anything similar to one of these, so multiply by $S$ on the left and $S^{-1}$ on the right, where $S$ is any invertible $2 \times 2$ matrix.
A: By the equation $\det(A)=\mathrm{trace}(A)$ so if $\lambda_1,\lambda_2$ are the eigenvalues of $A$ so
$$
\lambda_1\lambda_2=\lambda_1+\lambda_2
$$
Then
$$
\lambda_1=\frac{\lambda_2}{\lambda_2-1}
$$
So $\sigma(A)=\{t,\frac{t}{t-1} \}$ if $t\in\mathbb{R}-\{0,2\}$ $A$ will be diagonalisable and $A=PDP^{-1}$ where $P$ is any invertible matrix and $D=\mathrm{Diag}(t,\frac{t}{t-1})$
If $t=0$ or $t=2$ we can use the jordan normal form to write :
$$
A=PJ_0P^{-1} \qquad \qquad  \textrm{ Or } \qquad A=PJ_2P^{-1}
$$
A: Since any $n\times n$ matrix $A\in\mathbb{C}$ is unitarily similar to upper triangular matrix, i.e. $P^{-1}AP=D, \:\overline{P}^T=P^{-1}=P$, there is
$$
\det e^A=\det e^{PDP^{-1}}=\det{P^{}}\det e^{D}\det{P^{-1}}=e^{\sum_{i=1}^n\lambda_i}=e^{\operatorname{trace}A}
$$
While $\det A=\prod_{i=1}^n\lambda_i$, so 
$$
\det e^A=e^{\det A}\iff \prod_{i=1}^n\lambda_i=\sum_{i=1}^n\lambda_i
$$
