Find solution to system of differential equations with initial conditions
Apart from some confusing notation with indices $(1,2)$ - better replace that with e.g. $(a,b)$ -
you've got everything OK. About the initial conditions:
$$
\vec{x}(t) = c_1 e^t \begin{pmatrix}-1 \\ 1\end{pmatrix} + c_2 e^{-t} \begin{pmatrix}1 \\ 1\end{pmatrix}
\quad \Longrightarrow \quad
\vec{x}(0) = c_1 \begin{pmatrix}-1 \\ 1\end{pmatrix} + c_2 \begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix}-1 & 1
\\ 1 &1 \end{pmatrix}\begin{pmatrix}c_1 \\ c_2\end{pmatrix}
$$ $$
\begin{pmatrix}-1 & 1\\ 1 &1 \end{pmatrix}\begin{pmatrix}c_1 \\ c_2\end{pmatrix}
= \begin{pmatrix} x_1(0) \\ x_2(0) \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}
\quad \Longrightarrow \quad
\begin{pmatrix}c_1 \\ c_2\end{pmatrix} =
\begin{pmatrix}-1 & 1\\ 1 & 1 \end{pmatrix}^{-1} \begin{pmatrix} -1 \\ 1 \end{pmatrix}
$$
Inverse of an almost orthogonal matrix is almost the transpose, so:
$$
\begin{pmatrix}c_1 \\ c_2\end{pmatrix} = \frac{1}{2}
\begin{pmatrix}-1 & 1\\ 1 & 1 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix}
\quad \Longrightarrow \quad
\begin{pmatrix}c_1 \\ c_2\end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}
$$