Eigenvalue calculation. I am getting confused by this simple eigenvalue calculation.

Calculate the eigenvalues of $\begin{bmatrix} 5 & -2\\ 1 & 2\end{bmatrix}$.

Firstly, I row reduce it, to go from $\begin{bmatrix} 5 & -2\\ 1 & 2\end{bmatrix} \to \begin{bmatrix} 6 & 0\\ 1 & 2\end{bmatrix}$ by performing $R_1 \to R_1 + R_2$.
The resulting equation is $(6 - \lambda)(2 - \lambda) = 0$ so the eigenvalues should be $6$ and $2$, but I check on wolfram alpha and it says they are $4$ and $3$.
Can someone please explain?
 A: An eigenvalue of the $n\times n$ matrix $A$ is a number $\lambda$ such that there is a nonzero vector $v$ with
$$
Av=\lambda v.
$$
Depending on the context you may be looking for $\lambda$ in the real or complex numbers. In any case, the equation is equivalent to
$$
(A-\lambda I)v=0
$$
where $I$ is the $n\times n$ identity matrix. This has a nonzero solution if and only if the rank of $A-\lambda I$ is less than $n$ or, equivalently,
$$
\det(A-\lambda I)=0.
$$
It turns out that the expression $\det(A-\lambda I)$ is a polynomial in $\lambda$ of degree exactly $n$, called the characteristic polynomial of $A$. Its roots are precisely the eigenvalues of $A$.
It's not possible to use row-reduction for finding eigenvalues. Computing the roots of the characteristic polynomial is not the only way: in some special cases other methods are available. Nevertheless, this method is always available.
In your case
$$
A-\lambda I=
\begin{bmatrix}
5 & -2 \\
1 &  2
\end{bmatrix}
-
\lambda
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
$$
so the characteristic polynomial is
$$
\det(A-\lambda I)=\det
\begin{bmatrix}
5-\lambda & -2 \\
1 & 2 - \lambda
\end{bmatrix}
=(5-\lambda)(2-\lambda)+2=\lambda^2-7\lambda+12
$$
and the roots are easily computed to be $3$ and $4$.
A: $\det \left(\begin{bmatrix} 5 -\lambda & -2\\ 1 & 2-\lambda\end{bmatrix}\right)=0$
therefore,
$(5 - \lambda)(2 - \lambda) - (-2 \times 1) = 0$
$(5 - \lambda)(2 - \lambda) = -2$
Solving that gives $\lambda\in \{4,3\}$.
A: Characteristic equation of any $2\times 2$ matrix is given by $\alpha^{2}-S_{1}\alpha+S_{2}=0$
, where $S_{1}=$ sum of principle diagonal element and $S_{2}=$ determinant of given matrix, 
using this we can easily get eigen values of matrix.
So, here we have,
$$\alpha^{2}-7\alpha+12=0$$
$$\Rightarrow (\alpha-4)(\alpha-3)=0$$
$$\therefore \alpha =3, 4 .$$
