## Linear Algebra and Its Applications (5th Edition)

If $h = 2$, then the system has no solution, otherwise, when $h \neq 2$, the system has a solution.
We begin with matrix: $\begin{bmatrix} 1 & h & 4\\ 3 & 6 & 8 \end{bmatrix}$ First, we need to multiply the first row by $-3$ and add it to the second row: $\begin{bmatrix} 1 & h & 4\\ 0 & 6-3h & -4 \end{bmatrix}$ Write $c$ for $6 – 3h$. If $c=0$ then $6-3h=0$. The solution of this equation is $h=2$. If $c = 0$, that is, if $h = 2$, then the system has no solution, because $0$ cannot equal $–4$. Otherwise, when $h \neq 2$, the system has a solution.