Find the kernel of a 4x4 matrix $$
    \begin{pmatrix}
    1 & 2 & 3 & 4\\
    5 & 6 & 7 & 8\\
    9 & 10 & 11 & 12\\
    13 & 14 & 15 & 16\\
    \end{pmatrix}
$$
I am asked to find the kernel of the matrix $M$. After doing some row operation I get to 
$$
    \begin{pmatrix}
    1 & 2 & 3 & 4\\
    0 & -4 & -8 & -12\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    \end{pmatrix}
$$
and for $x$ I find $x = \alpha + 2\beta$, whereas $y = -2\alpha -3\beta$
Therefore, 
$$
    \begin{pmatrix}
    \alpha + 2\beta\\
    -2\alpha - 3\beta\\
    \alpha\\
    \beta\\
    \end{pmatrix}
$$
When we take outside alpha and beta:
we get two vectors:
$$
    \begin{pmatrix}
    1\\
    -2\\
    1\\
    0\\
    \end{pmatrix}
$$
and
$$
    \begin{pmatrix}
    2\\
    -3\\
    0\\
    1\\
    \end{pmatrix}
$$
which are linearly independent and form a basis of this $ker(M)$
Could you please confirm with me whether you get the same result? Thank you.

Yes it is correct, in case of doubt you can check it directly by simple multiplication for RREF matrix
$$\begin{bmatrix}
    1 & 2 & 3 & 4\\
    0 & -4 & -8 & -12\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    \end{bmatrix}\begin{bmatrix}
    1\\
    -2\\
    1\\
    0\\
    \end{bmatrix}=0$$
$$\begin{bmatrix}
    1 & 2 & 3 & 4\\
    0 & -4 & -8 & -12\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    \end{bmatrix}\begin{bmatrix}
    2\\
    -3\\
    0\\
    1\\
    \end{bmatrix}
=0
$$
or/and also for the original matrix
$$\begin{bmatrix}
    1 & 2 & 3 & 4\\
    5 & 6 & 7 & 8\\
    9 & 10 & 11 & 12\\
    13 & 14 & 15 & 16\\
    \end{bmatrix}
\begin{bmatrix}
    1\\
    -2\\
    1\\
    0\\
    \end{bmatrix}=0$$
$$\begin{bmatrix}
    1 & 2 & 3 & 4\\
    5 & 6 & 7 & 8\\
    9 & 10 & 11 & 12\\
    13 & 14 & 15 & 16\\
    \end{bmatrix}
\begin{bmatrix}
    2\\
    -3\\
    0\\
    1\\
    \end{bmatrix}
=0
$$
Note that also after this check, for the correctness of the result it is crucial that RREF is calculated properly. In case of doubt on it we can evaluate $rank(M)$ with others methods or follow the nice suggestion give by Atmos to be sure that $dim(Ker M)=2$ and thus that what we have found is a basis for $Ker(M)$.
