Determine if a system of equations is independent, dependent or inconsistent Is there a way to determine the nature of a system of equations without solving it? For example, given the system 
\begin{cases}
    2x + y - 4z = 6 \\[4px]
    y - 2z = 2 \\[4px]
    4x + 3y - 10z = -3
\end{cases}
Can I tell that this system is independent without solving it?

Not really. You can see whether the three equations are independent by computing the determinant
$$
\det\begin{bmatrix}
2 & 1 & -4 \\
0 & 1 & -2 \\
4 & 3 & -10
\end{bmatrix}
$$
but, unfortunately, it is $0$. Besides, computing a determinant with Laplace expansion is a very expensive computation.
Much simpler is going with Gaussian elimination:
\begin{align}
\begin{bmatrix}
2 & 1 & -4 & 6\\
0 & 1 & -2 & 2\\
4 & 3 & -10 & -3
\end{bmatrix}
&\to
\begin{bmatrix}
2 & 1 & -4 & 6\\
0 & 1 & -2 & 2\\
0 & 1 & -2 & -15
\end{bmatrix}
&& R_3\gets R_3-2R_1
\\[6px]
&\to
\begin{bmatrix}
2 & 1 & -4 & 6\\
0 & 1 & -2 & 2\\
0 & 0 & 0 & -17
\end{bmatrix}
&& R_3\gets R_3-R_2
\end{align}
The last column is a pivot column, so the system is inconsistent.

The system would be solvable if $-10$ is changed into $14$. In this case we could go backwards
\begin{align}
\begin{bmatrix}
2 & 1 & -4 & 6\\
0 & 1 & -2 & 2\\
0 & 0 & 0 & 0
\end{bmatrix}
&\to
\begin{bmatrix}
2 & 0 & -2 & 4\\
0 & 1 & -2 & 2\\
0 & 0 & 0 & 0
\end{bmatrix}
&& R_1\gets R_1-R_2
\\[6px]
&\to
\begin{bmatrix}
1 & 0 & -1 & 2\\
0 & 1 & -2 & 2\\
0 & 0 & 0 & 0
\end{bmatrix}
&& R_1\gets\tfrac{1}{2}R_1
\end{align}
The third unknown is free, and the solutions are
\begin{cases}
x=2+h \\[4px]
y=2+2h \\[4px]
z=h
\end{cases}
with arbitrary $h$.
