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.
 A: 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 $Ker(M)$.
A: Yes it looks good great. However, you should precise why the dimension of Kernel is $2$. For example by extracting the $2 \times 2$ determinant
$$
\begin{vmatrix}
 3&4 \\ 
 7&8 
\end{vmatrix}=24-28 \ne 0
$$
So the dimesion of the image is at least $2$. Then you found two vectors in it, so the dimension of the image cannot exceed two so it values $2$. as well as the dimension of the Kernel.
A: Your computations are correct and so is your answer. But you should really improve your MathJax skills.
