Calculation of Cokernel? Is my solution correct? I was trying to find the following: 
$$ M= \begin{bmatrix}
    2 & 4\\
    3 & 2\\
    1 & 1 
\end{bmatrix}\\
coker(M) = (x,y,z)* 
\begin{bmatrix}
    2 & 4\\
    3 & 2\\
    1 & 1 
\end{bmatrix}= 0\\
2x+3y+z = 0\\
4x+2y+z = 0\\
z = -x+2y\\
2x+3y+(-4x-2y)=-2x+y \Rightarrow y = 2x \Rightarrow z = -8x\\
\Rightarrow coker(M) = \begin{bmatrix}x\\2x\\-8x \end{bmatrix}$$
Is that correct? 
 A: Your question is imprecise.
(1) Cokernels are defined for linear maps $f : V \to W$ between vector spaces $V, W$ over a field $K$.
You do not say which field is considered and which are the vector spaces. Most likely you have $K = \mathbb{R}$. Then I would conclude that $V = \mathbb{R}^3, W = \mathbb{R}^2$. See the next point.
(2) You consider a matrix $M$, not a linear map.
Of course, $M$ is the matrix representation of a unique linear map $f : \mathbb{R}^3 \to \mathbb{R}^2$ with respect to the standard bases of $\mathbb{R}^3$ and $\mathbb{R}^2$. Writing the elements of $\mathbb{R}^n$ as row vectors, you have $f(x) = x * M$, where $*$ denotes matrix multiplication.
(3) The cokernel of $f$ is defined as the quotient space $W / \text{im}(f)$. It seems you consider the kernel of $f$ which defined as $\text{ker}(f) = \{ x \in V \mid f(x) = 0 \}$. In our case $\text{ker}(f) = \{ x \in V \mid x * M  = 0 \}$. This was calculated correctly.
Perhaps I misunderstood something, but you see the some more precision would be needed. I encourage you to do this in future questions.
