Show that the Lie algebra generated by the given matrices is isomorphic to $\mathfrak{so}(3)$ I'm trying to show that the Lie algebra $\cal{L}$
generated by the matrices
\begin{equation}  
t_1 =  
\begin{pmatrix} 
0 &  0 &  0 \\ 
0 &  0 & -1 \\ 
0 & +1 &  0 
\end{pmatrix} 
\quad 
t_2 =  
\begin{pmatrix} 
 0 & 0 & x \\ 
 0 & 0 & 0 \\ 
-1 & 0 & 0 
\end{pmatrix} 
\quad
t_3 =  
\begin{pmatrix} 
 0 & -x & 0 \\ 
+1 &  0 & 0 \\ 
 0 &  0 & 0 
\end{pmatrix}\\ 
\end{equation}
with 
$x \in \mathbb{R}$ and $x \ne 1$
is isomorphic to
$\mathfrak{so}(3)$.
The corresponding commutation relations are
\begin{equation}  
  [t_1, t_2] = t_3 
  \quad
  [t_2, t_3] = x \, t_1 
  \quad
  [t_3, t_1] = t_2
\end{equation}
Specifically, I'm looking for a matrix $A$
such that
\begin{equation}  
  \epsilon_{ijk}=\sum_{n,m,s} A_{in}\,A_{jm}\,(A^{-1})_{sk}\,f_{nms}
\end{equation}
with $f_{nms}$ denoting ${\cal L}$'s structure constants 
\begin{equation}  
  f_{2,3,1} = x = -f_{3,2,1} \qquad
  f_{3,1,2} = f_{1,2,3} = 1 = -f_{1,3,2} = -f_{2,1,3}
\end{equation}
and the 
Levi-Civita symbol 
$\epsilon_{ijk}$, 
the structure constants of $\mathfrak{so}(3)$.
How do I construct $A$?

Hint: try correcting the relations rather than looking for an explicit isomorphism. More specifically, the first and last relation (which are the same as for $\mathfrak{so}(3)$) remain unchanged if you multiply $t_2$ and $t_3$ by the same factor. Now choose that factor in such a way that you "correct" the middle relation. 
