Conditions for consistent matrix

Question

In my professor's notes he said that the matrix

$$ \left[\begin{array}{rr|r} 1 & -2 & 4 \\ 0 & 5 & -15 \\ 0 & 3 & h+8 \end{array}\right]$$

is consistent when $3=h+8$. Why is that? It seems that if that is true, then $x_2=1$ which means that $5x_2=-15$ can't be true. If this is a mistake of his (which he has been known to make) what are the values of $h$ which make this matrix consistent?

Answer 1

In order to be consistent, the system of linear equations must have a solution. Given $$x_1−2x_2=4\tag{1}$$ $$5x_2=−15\tag{2}$$ $$3x_2=h+8\tag{3}$$ we can solve $(2)$ to find $x_2=-3$. Substituting $x_2=-3$ into $(1)$ produces $x_1=-2$. Then, substituting $x_2=-3$ into $(3)$ forms $$-9=h+8 \implies h=-17$$ so the solution is given by $$x_1=-2, ~x_2=-3,~h=-17$$ Your professor needed to write $-9=h+8$.

Answer 2

From the second line, $x_2=-3$.

From the last line, $3(-3) = h+8$, that is you are right, there is a mistake.