Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix. 
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4  \end{array}\right) \in
M_n(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.

How can I find $P$? I am doing Gauss but it does not work?$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1    \end{array}\right) \sim \left(\begin{array}{cc|cc} 2&0&-8&6\\ 0&-1/2&-3/2&1    \end{array}\right)$$
What am I doing wrong? Steps would be much appreciated.

Problem
Diagonalize the matrix
$$
\mathbf{A} =
\left[
\begin{array}{cc}
 2 & 3 \\
 3 & 4 \\
\end{array}
\right]
$$

Solution
Compute eigenvalues
The eigenvalues are the roots of the characteristic polynomial
$$
 p(\lambda) = \lambda^{2} - \lambda \text{ trace }\mathbf{A} + \det \mathbf{A}
$$
The trace and determinant are
$$
\text{ trace }\mathbf{A} = 6, \qquad \det \mathbf{A} = -1
$$
Therefore
$$
 p(\lambda) 
= \lambda^{2} - \lambda \text{ trace }\mathbf{A} + \det \mathbf{A}
= \lambda^{2} - 6 \lambda - 1
$$
The roots are the eigenvalue spectrum
$$
 \lambda \left( \mathbf{A} \right) = 3 \pm \sqrt{10}
$$
Result:
$$
\mathbf{D} = \left[
\begin{array}{cc}
 3+\sqrt{10} & 0 \\
 0 & 3-\sqrt{10} \\
\end{array}
\right]
$$
Eigenvectors
First
$$
\begin{align}
  \left(\mathbf{A} - \lambda_{1} \mathbf{I}_{2} \right) w_{1} &= \mathbf{0} \\
%
  \left[
\begin{array}{cc}
 -1-\sqrt{10}-1 & 3 \\
 3 & 1-\sqrt{10} \\
\end{array}
\right]
%
  \left[
\begin{array}{c}
 w_{x} \\
 w_{y} \\
\end{array}
\right]
%
&=
%
  \left[
\begin{array}{c}
 0 \\
 0 \\
\end{array}
\right]
%
\end{align}
$$
Solution
$$
 w_{1} = 
  \left[
\begin{array}{c}
 \frac{1}{3} \left(-1+\sqrt{10}\right) \\
 1 \\
\end{array}
\right]
$$
Second
$$
\begin{align}
  \left(\mathbf{A} - \lambda_{2} \mathbf{I}_{2} \right) w_{2} &= \mathbf{0} \\
%
  \left[
\begin{array}{cc}
 -1+\sqrt{10} & 3 \\
 3 & 1+\sqrt{10} \\
\end{array}
\right]
%
  \left[
\begin{array}{c}
 w_{x} \\
 w_{y} \\
\end{array}
\right]
%
&=
%
  \left[
\begin{array}{c}
 0 \\
 0 \\
\end{array}
\right]
%
\end{align}
$$
Solution
$$
 w_{2} = 
  \left[
\begin{array}{c}
 -\frac{1}{3} \left(1+\sqrt{10}\right) \\
 1 \\
\end{array}
\right]
$$
Diagonalization matrix
$$
 \mathbf{P} =
\left[
\begin{array}{cc}
 \frac{1}{3} \left(-1+\sqrt{10}\right) & -\frac{1}{3} \left(1+\sqrt{10} \right) \\
 1 & 1 \\
\end{array}
\right],
\qquad 
\mathbf{P}^{-1} =
\frac{1}{2\sqrt{10}}
\left[
\begin{array}{rr}
 3 & 1+\sqrt{10} \\
 -3 & -1+\sqrt{10} \\
\end{array}
\right]
$$

Validation
You can check that
$$
 \mathbf{P} \mathbf{A} \mathbf{P}^{-1} = \mathbf{D}
$$
and
$$
 \mathbf{P}^{-1} \mathbf{D} \mathbf{P} = \mathbf{A}
$$


Gaussian elimination
Solving this problem does not require Gaussian elimination. However, since you specifically asked, here is the process:
Clear column 1
$$
 \left[
\begin{array}{rc}
  \frac{1}{2} & 0 \\
 -\frac{3}{2} & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{cc|cc}
 2 & 3 & 1 & 0 \\
 3 & 4 & 0 & 1 \\
\end{array}
\right]
=
\left[
\begin{array}{cr|rc}
 1 &  \frac{3}{2} &  \frac{1}{2} & 0 \\
 0 & -\frac{1}{2} & -\frac{3}{2} & 1 \\
\end{array}
\right]
$$
Clear column 2
$$
 \left[
\begin{array}{cr}
 1 & 3 \\
 0 & -2 \\
\end{array}
\right]
%
\left[
\begin{array}{cr|rc}
 1 &  \frac{3}{2} &  \frac{1}{2} & 0 \\
 0 & -\frac{1}{2} & -\frac{3}{2} & 1 \\
\end{array}
\right]
=
\left[
\begin{array}{cc|rr}
 1 & 0 & -4 & 3 \\
 0 & 1 & 3 & -2 \\
\end{array}
\right]
$$
