Consistent but apparently unsolvable system of equations I have started learning linear algebra. One of the exercise problems is as below.
Determine the value(s) of $h$ such that the matrix is the augmented matrix of a consistent linear system.
$$
\left [
        \begin{matrix}
        1 & 4 &| -2 \\
        3 & h &| -6 \\
        \end{matrix}
\right ]
$$
A simple row transformation of $R_2 = R_2 - 3R_1$ results in 
$$
\left [
        \begin{matrix}
        1 & 4 &| -2 \\
        0 & h-12 &|\ \ \ \  0 \\
        \end{matrix}
\right ]
$$
I reasoned that if $h$ is $12$, though there wouldn’t be any inconsistency, the system would effectively have only one equation to solve for two unknown variables and so $h \neq 12$ if we are to find the solution.
Generally how do we classify such systems, the ones which are consistent but appear unsolvable? On the other hand, is my reasoning wrong?
 A: Your reasoning is correct. At $h = 12$, the system is consistent. 
Such a system, in your case, when $h = 12$, or in general, when the augmented coefficient matrix of a system of linear equations has a row of zeros in its row-reduction, we say that the corresponding system has an infinite number of solutions.

Let's look at the system represented by your matrix. 
Essentially, we can choose any arbitrary value for $y$: put $y = \alpha$.
Then $x = -2 - 4y = -2(1 + 2y) = -2(1 + 2\alpha)$
We can then represent the infinite "family" of solutions by the vector:
$$\begin{pmatrix}x \\y\end{pmatrix}:=\begin{pmatrix}-2(1 + 2\alpha) \\\alpha \end{pmatrix}$$
For any chosen value (scalar) $\alpha$, $x$ is then determined, so our solution puts some restrictions on the possible solutions, even though there are infinitely many choices for $\alpha$. 
So the system having an infinite number of solutions is not equivalent to saying that every vector $\begin{pmatrix} x \\ y \end{pmatrix}$ is a solution to the system, since $x$ is a function of $y$.
