Explain why this augmented matrix is consistent for all h? So I'm attempting to study for an exam tomorrow and for the life of me, cannot figure out why this matrix is consistent for all h (instead of being only consistent for h=10).
The matrix is:
$\begin{bmatrix}
1 & 2 & -4\\ 
5 & h & -20
\end{bmatrix}$
I've row reduced it to the form:
\begin{bmatrix}
1 & 2 & -4\\ 
0 & -10+h & 0
\end{bmatrix}
but the reasoning eludes me, could someone explain?
 A: the system $Ax = b$ is inconsistent if the augmented matrix $A|b$ has a pivot on the last column. it is because you get an equation of the form $0x = 1.$ in your case the reduced from of the augmented matrix cannot have a pivot in the last column. therefore $Ax = b$ is consistent for any $h.$
A: You were able to see that
$$
\begin{bmatrix}
1 & 2 & -4\\ 
5 & h & -20
\end{bmatrix}
\sim
\begin{bmatrix}
1 & 2 & -4\\ 
0 & -10+h & 0
\end{bmatrix}
$$
but what does this really mean? To get a better grasp of the situation, it might help you to look at the matrices in terms of what they actually represent, i.e. a system of two linear equations. The right-hand matrix you obtained by a row reduction represents the following linear equations (top row and bottom row, respectively):
$$
x_1 +2x_2 = -4\tag{1}
$$
and 
$$
0x_1+(-10+h)x_2 = 0.\tag{2}
$$
When $h=10$, we see that $(2)$ amounts to 
$$
0x_1+0x_2=0,
$$
and this is obviously true for all values of $x_1$ and $x_2$, and thus the system is consistent when $h=10$. What if $h\neq 10$ though? Can the system still be consistent?
If $h\neq 10$, then we have $h-10=b\neq 0$, and we can represent the row-reduced matrix as
\begin{bmatrix}
1 & 2 & -4\\ 
0 & b & 0
\end{bmatrix}
Notice what this means in the context of $(2)$:
$$
0x_1+bx_2=0 \Longleftrightarrow x_2=\frac{0}{b}.
$$
This simplifies your work quite a bit, for now you are dealing with the two linear equations $x_1+2x_2=-4$ and $x_2=\frac{0}{b}$. Substitute $x_2=\frac{0}{b}$ into the first linear equation and see what happens:
$$
x_1+2\left(\frac{0}{b}\right)=-4.
$$
With $x_1=-4$ this equation will be true for any value of $b=h-10$ or, equivalently, any value of $h$. Thus, your system of linear equations will be consistent regardless of your choice for $h$. That's a pretty dirty explanation, but hopefully it will give you a more intuitive grasp of the situation.
A: If $\;h=10\;$ then the second row is a multiple of the first one and thus the system reduces to
$$x+2y=-4$$
and this  thing always has solution.
