0
$\begingroup$

I don't understand how this statement is FALSE. What if a matrix resulted in a row which led us to row 0x2 = 9, which would tell us that the plane or vector is parallel?

Thanks in advance for clearing up my confusion.

Reference : This was from my linear algebra textbook. Elementary Linear Algebra Tenth Edition by Howard Anton and Chris Rorres.

Chapter 1.1 True False exercise (e)

$\endgroup$
  • 1
    $\begingroup$ The statement in the title is false. $\endgroup$ – user296602 Jan 2 '16 at 20:53
  • $\begingroup$ This was from my textbook, I added the reference. Yea I think it may be wrong as well.. $\endgroup$ – Mohit A. Jan 2 '16 at 20:56
  • $\begingroup$ The system $x+y+z=1,\ x+y+z=2$ is inconsistent. And what do you mean by "a matrix resulted in a row which led us to row 0x2 = 9": do you mean during Gaussian elimination? $\endgroup$ – Rory Daulton Jan 2 '16 at 20:57
  • $\begingroup$ Am i not reading the question properly? Let me change the title to match exactly what the book is saying $\endgroup$ – Mohit A. Jan 2 '16 at 20:58
  • $\begingroup$ Yes, by Gaussian Elimination. The example you provided we have more variables than equations, the question is asking for more equations than variables $\endgroup$ – Mohit A. Jan 2 '16 at 21:04
3
$\begingroup$

The key word here is must. I.e., the statement claims that every system of linear equations with more equations than unknowns is inconsistent. That’s false. For example, the system $$\begin{align} x &= 1 \\ 2x &= 2 \end{align}$$ has two equations and one unknown, but is clearly consistent.

$\endgroup$
  • $\begingroup$ Oh man, that definitely works. Wow what a way to confuse what they are trying to say. Thanks! $\endgroup$ – Mohit A. Jan 2 '16 at 22:23
0
$\begingroup$

The statement in the title isn't necessarily true. It's entirely possible for a system to be inconsistent even though the number of equations is greater than the number of variable.

Maybe the statement was taken out of context?

$\endgroup$
  • $\begingroup$ I changed the title, to exactly what the question says. $\endgroup$ – Mohit A. Jan 2 '16 at 21:14

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.