Find real $P$ s.t. $B=P^{-1}AP$ TASK:
$\frac{dx}{dt}=Ax$
Given $A=\begin{bmatrix}
    1     & 0 & 0 \\
    0       & 2 & -3 \\
    1      & 3 &2
\end{bmatrix}$
Find real matrix $P$ s.t change of coordinates $x=Py$ transforms the system to
and 
$\frac{dy}{dt}=By$
$B=\begin{bmatrix}
    1     & 0 & 0 \\
    0       & 2 & -3 \\
    0      & 3 &2
\end{bmatrix}$
Solve explicitly for $y$ hence evaluate solution in terms of $x$
APPROACH:
Firstly $PB=AP$
I tried to solve the 9 simultanouesequations and ended up with matrix where there are 3 free variables of the form:
$P=\begin{bmatrix}
    -10x     & 0 & 0 \\
    3x       & y & -z \\
    x      & z &y
\end{bmatrix}$
Then I put this into $P^{-1}AP$ in software "SYMBOLAB" and the asnwer he gave me was $B$. So it turns out $x,y,z$may are free as long as $det(P)\neq0$ And since $det(P)=-10x(y^2+z^2)$, all I know is that $x\neq 0$ and at least one of the $y,z$ is not 0.
Is this the correct answer? 

Besides the hint given in my comment to a related post, consider that the similarity relation is not unique, because in fact:
$$
\mathbf{A = R}\;\mathbf{C}\;\mathbf{R}^{\,\mathbf{ - 1}} \quad \quad  \Leftrightarrow \quad \left\{ {\begin{array}{*{20}c}
   {\mathbf{A} = \mathbf{A}\;\mathbf{A}\;\mathbf{A}^{\,\mathbf{ - 1}} } &  \Leftrightarrow  & {\mathbf{A = }\left( {\mathbf{A}\;\mathbf{R}} \right)\;\mathbf{C}\;\left( {\mathbf{A}\;\mathbf{R}} \right)^{\,\mathbf{ - 1}} }  \\
   {\mathbf{C} = \mathbf{C}\;\mathbf{C}\;\mathbf{C}^{\,\mathbf{ - 1}} } &  \Leftrightarrow  & {\mathbf{A = }\left( {\mathbf{R}\;\mathbf{C}} \right)\;\mathbf{C}\;\left( {\mathbf{R}\;\mathbf{C}} \right)^{\,\mathbf{ - 1}} }  \\
   \begin{gathered}
  \mathbf{A} = \lambda \;\mathbf{A}\;\lambda ^{\, - 1}  \hfill \\
  \mathbf{C} = \mu \;\mathbf{C}\;\mu ^{\, - 1}  \hfill \\ 
\end{gathered}  &  \Leftrightarrow  & {\mathbf{A = }\left( {\lambda \;\mathbf{R}\;\mu } \right)\;\mathbf{C}\;\left( {\lambda \;\mathbf{R}\;\mu } \right)^{\,\mathbf{ - 1}} }  \\
 \end{array} } \right.
$$
