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 $-4(x_1-x_2/2)=2 \Leftrightarrow x-1-x_2/2 = -1/2$


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