Solve $\dot X = AX$ and find the time at which the area doubles. We have a linear operator $X(t): \mathbb{E}^2 \to \mathbb{E}^2$ such that
$$\dot X = AX\quad X(0)=\mathbf{1}$$
To be clear $X$ is a $2\times 2$-matrix. An ink spot is contained in $\mathbb{E}^2$ at $t=0$. At which time $t$, will the area of the image of the ink spot under $X(t)$, double for $A=\left(\begin{matrix}-1 & 0\\ 0 & -1\end{matrix}\right)$?

I know the solution is $X(t) = \exp(tA)$. Now the area of a scaling $2\times  2$ matrix is the absolute value of the determinant. So we have
$$|\det(X(t))|= |\det(\exp(tA))|= |\exp(tr(tA))|= \exp(-2t)$$
For $t=0$ we have $1$. So we need to solve 
$\exp(-2t) = 2$ that is $t= \frac{\ln 2}{-2}$. 
Is this solution correct?

I know the solution is $X(t) = \exp(tA)$. Now the area of a scaling $2\times  2$ matrix is the absolute value of the determinant. So we have
$$|\det(X(t))|= |\det(\exp(tA))|= |\exp(tr(tA))|= \exp(-2t)$$
For $t=0$ we have $1$. So we need to solve 
$\exp(-2t) = 2$ that is $t= \frac{\ln 2}{-2}$. 
Posting my working as an answer to close this question. 
