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

The system is consistent for all $h$.
We begin with matrix $\begin{bmatrix} 1 &amp; h &amp; -3 \\ -2 &amp; 4 &amp; 6 \\ \end{bmatrix}$ First, we need to multiply the first row by $2$ and add it to the second row: $\begin{bmatrix} 1 &amp; h &amp; -3 \\ 0 &amp; 2h+4 &amp; 0 \\ \end{bmatrix}$ Write $c$ for $2h+4$. Then the second equation $c\cdot x_2 = 0$ has a solution for every value of $c$. So the system is consistent for all $h$.