How find the matrix $X$ such $e^{X}$ is give it 
Question:
let matrix $$X=\begin{bmatrix}
a&b\\
c&d
\end{bmatrix} ,e^{X}=\begin{bmatrix}
-1&2\\
0&-1
\end{bmatrix}$$
  and such $a+d=0$,
Find the matrix $X$
  my idea
  $$e^{Tr{(X)}}=det{(e^{X})}$$
  so
  $$1=1$$ is true

I think we can find the matrix function: $e^X$,but I have find this is ver ugly,so I think this problem have other methods
 A: Firstly, since the eigenvalues of $e^X$ are the exponentials of the eigenvalues of $X$, $X$ can only have eigenvalues of the form $i(\pi+2n\pi)$. Since $e^X$ is not diagonalizable, neither is $X$, so $X$ only has one eigenvalue, say $i\pi$. So we can try $X=\begin{bmatrix} i\pi&z\\ 0&i\pi\end{bmatrix}$. Then, $e^X=e^{i\pi I}e^{\begin{bmatrix} 0&z\\0&0\end{bmatrix}}=-I\left(I+\begin{bmatrix} 0&z\\ 0&0\end{bmatrix}\right)=\begin{bmatrix} -1&-z\\ 0&-1\end{bmatrix}$. So, we can let $z=-2$, and then we see that $\exp\left({\begin{bmatrix} i\pi&-2\\ 0&i\pi\end{bmatrix}}\right)=\begin{bmatrix} -1&2\\ 0&-1\end{bmatrix}$.  
A: Here's a sketch. Since $d = -a$, you can write $X = \left(\begin{array}{rr} a & b \\ c & -a\end{array}\right)$, which gives
$$X^2 = \left(\begin{array}{cc} a^2 + bc & 0 \\ 0 & a^2 + bc\end{array}\right) = (a^2+bc)I.$$
This allows you to write a general expression for $X^n$: $X^{2n} = (a^2+bc)^n I$ and likewise $X^{2n+1} = (a^2+bc)^nX$.
With these two in hand, we can evaluate the exponential:
$$\exp X = \sum_{n=0}^{\infty} \frac{1}{n!} X^n = \sum_{n=0}^{\infty} \frac{1}{(2n)!} X^{2n} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}X^{2n+1}.$$
Making use of our above observation:
$$\exp X = \sum_{n=0}^{\infty} \frac{(a^2+bc)^n}{(2n)!}\, I+ \sum_{n=0}^{\infty} \frac{(a^2+bc)^n}{(2n+1)!}\,X.$$
Since $\exp X = \left(\begin{array}{rr} -1 & 2 \\ 0 & -1\end{array}\right)$, we see that $c= 0$, so that
$$\exp X = \sum_{n=0}^{\infty} \frac{a^{2n}}{(2n)!}\, I+\sum_{n=0}^{\infty} \frac{a^{2n}}{(2n+1)!}\,X.$$
From here, if you equate terms, you will find that $a = i\pi+2k\pi$ for some $k\in\Bbb Z$ and $b = -2$ as Nishant derived.
