I have been a bit confused about this linear algebra question, if someone can explain it would be great. So my professor is asking us to determine all values of x for which the linear system a, is consistent b, is inconsistent:
$$\left(\begin{array}{cc|c} 1 & h & 1\\ -4 & 2 & 2\\ \end{array}\right)$$
They are all one! just didnt know how to put big braces! I have simplified a bit and have gotten
$$\left(\begin{array}{cc|c} 1 & h & 1\\ 0 & h & 3/2\\ \end{array}\right)$$
can someone explain what should be the next step? Thank you
I think you've made a mistake to start with, your simplification doesn't seem to be correct. You start with
$$\begin{pmatrix}1 & h \\ -4 & 2\end{pmatrix}x = \begin{pmatrix}1 \\ 2\end{pmatrix}$$
Then to simplify it you add four times the first row to the second
$$\begin{pmatrix}1 & h \\ 0 & 2+4h\end{pmatrix}x = \begin{pmatrix}1 \\ 6\end{pmatrix}$$
And then we add $-h/(2+4h)$ times the second line to the first:
$$\begin{pmatrix}1 & 0 \\ 0 & 2+4h\end{pmatrix}x = \begin{pmatrix}1 - 6h/(2+4h) \\ 6\end{pmatrix}$$
This looks consistent, but we have to be sure that $2+4h\ne0$ for the above to work out. If we had $h=-1/2$ we would have
$$\begin{pmatrix}1 & -1/2 \\ -4 & 2\end{pmatrix}x = \begin{pmatrix}1 \\ 2\end{pmatrix}$$
Now we see that the second row of the LHS is $-4$ times the first, but that's not the case on the RHS. So for $h=-1/2$ we have an inconsistency: the first line says $x_1-x_2/2 = 1$ but the second says $-4(x_1-x_2/2)=2 \Leftrightarrow x-1-x_2/2 = -1/2$