Linear algebra augmented matrix $$
        \begin{bmatrix}
        2 & h & 4 \\
        3 & 6 & 7 \\
        \end{bmatrix}
$$
Find $h$ such that this matrix is inconsistent 
what I have done:
switched $R_1$ and $R_2$
$$
        \begin{bmatrix}
        3 & 6 & 7 \\
        2 & h & 4 \\
        \end{bmatrix}
$$
$R_2$ replaced by $(-3)R_2 + 2(R_1)$
$$
        \begin{bmatrix}
        3 & 6 & 7 \\
        0 & 12-3h & 2 \\
        \end{bmatrix}
$$
$R_2/(12-3h)$
$$
        \begin{bmatrix}
        3 & 6 & 7 \\
        0 & 1 & 2/(12-3h) \\
        \end{bmatrix}
$$
$R_1$ replaced by $(-6)R_2 + R_1$
$$
        \begin{bmatrix}
        3 & 0 & (4/h+4)+7 \\
        0 & 1 & 2/(12-3h) \\
        \end{bmatrix}
$$
$R_1/3$
$$
        \begin{bmatrix}
        1 & 0 & (4/(3(h+4))+7/3 \\
        0 & 1 & 2/(12-3h) \\
        \end{bmatrix}
$$
I feel like I am doing this right but I followed along on another example on the website and got lost near the end, can someone tell me where I went wrong?
 A: I'm sure you know that you must never divide by zero.  But, even more important than this, never divide by something that might be zero.
Dividing by $12-3h$ is wrong because you do not know whether or not it is zero - after all, $h$ is an unknown, that is the whole point of the question.
Also, there is no need to reduce a matrix all the way to fully-reduced form (unless your instructor insists that you always do this - which IMHO would be very foolish). 
The best solution is to stop after your second matrix.  This will be inconsistent if the left hand side is zero and the right hand side is not, which occurs if and only if $h=4$.
In fact, there is really no need even for your first step: you could just do $2(R_2)-3(R_1)$ in the original matrix and you would reach the same conclusion.
A: Another way to do this is to note that your system is $A\vec x=\vec b$ where
\begin{align*}
A&=
\left[\begin{array}{rr}
2 & h \\
3 & 6
\end{array}\right] &
\vec b &=
\begin{bmatrix}
4\\ 7
\end{bmatrix}
\end{align*}
Since $\det A=-3 \, h + 12$ we know that $A\vec x=\vec b$ has a solution if $h\neq 4$.
So, it remains to determine if the system given by
$$
\left[\begin{array}{rr|r}
2 & 4 & 4 \\
3 & 6 & 7
\end{array}\right]
$$
Row reducing gives
$$
\DeclareMathOperator{rref}{rref}\rref
\left[\begin{array}{rr|r}
2 & 4 & 4 \\
3 & 6 & 7
\end{array}\right]=
\left[\begin{array}{rr|r}
1 & 2 & 0 \\
0 & 0 & 1
\end{array}\right]
$$
Thus we see that the system is inconsistent when $h=4$.
