0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.