Below I will prove that the matrix $A$ is not invertible, using the linear transformation $T:V\to W$ that corresponds to $L_A$ (over a field $\mathbb{F}$).
The reason for posting is proof verification, and sharing a bit of knowledge in case it already is correct. I couldn't really find this method to check invertibility of matrices anywhere else; mostly, people refer to the determinant or just Gaussian elimination (without the isomorphism/bijection part which to me is an interesting insight).
Proof:
Let $A = \begin{pmatrix} 6 & 0 & 0 \\ 1 & 2 & 1 \\ 2 & 8 & 4 \end{pmatrix}$, then we can see that $T:\mathbb{F}^3\to \mathbb{F}^3$ is the linear transformation that corresponds to $L_A$, and that a row echelon form of $A$ is $\begin{pmatrix} 6 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 0 \end{pmatrix}$.
We know that $dim(V)=dim(W)=3$ and that $dim(V)=rank(T)+null(T)$. From the row echelon form of $A$ we can see that $rank(A)=rank(T)=2$ so that $dim(V)=2+null(T).$
However, for the matrix to be invertible, we need the linear transformation be bijective.
If $dim(V)=dim(W)=3$ and $rank(A)=2$ we see that $rank(T)\neq dim(W)\implies R(T)\neq W$ so the mapping is not onto and thus not bijective.
We also see that $null(T) = 3-2=1\neq0$ so the mapping is not one-to-one and thus not bijective.
In either case, we see that the matrix $A$ is not invertible.