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?

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 $h=4$.
