0
$\begingroup$

[Please critique my reasoning]

$$x_{1}-2x_{2}+3x_{3} = 2$$ $$x_{1}+x_{2}+x_{3} = c$$ $$2x_{1}-x_{2}+4x_{3} = c^{2}.$$

We obtain the augmented matrix, $$\left[\begin{array}{@{}ccc|c@{}} 1 & -2 & 3 & 2 \\ 1 & 1 & 1 & c \\ 2 & -1 & 4 & c^2 \end{array}\right]$$

and its RREF

$$\left[\begin{array}{@{}ccc|c@{}} 1 & 0 & \frac{5}{3} & \frac{2}{3}c + \frac{2}{3} \\ 0 & 1 & -\frac{2}{3} & \frac{c-2}{3} \\ 0 & 0 & 0 & c^2 - c \end{array}\right].$$

Therefore the system will have no solutions for any $c \neq 0, c \neq 1$ (the last row is of the form $[0 0 0 | k], k \neq 0$). Since $x_{3}$ is a free variable, the system cannot have a unique solution. It also follows that the system has infinite solutions whenever $c = 0$ or $c = 1$ (the last row is of the form $[0 0 0 | 0]$)

$\endgroup$
2
  • 1
    $\begingroup$ the last line right hand side is not $c^2-c,$ it is $c^2 - c - 2$ $\endgroup$
    – Will Jagy
    May 5 at 18:43
  • $\begingroup$ always worth checking with actual numbers; in this case, check $c=0,1$ in the original system. We make fewer errors with numbers than with symbols $\endgroup$
    – Will Jagy
    May 5 at 20:59
1
$\begingroup$

Yes, your reasoning and answer are fine. (Assuming you've calculated the RREF correctly.)

Edit: Will Jaggy points out that it should be $c^2 - c - 2$ in the RREF, instead of $c^2 - c$. In that case, instead of $c \neq 0, 1$, you'll modify that to $c \neq 2, -1$. The remaining goes as it is.


In fact, you can note the following for systems of the form $Ax = b$ where $A \in \Bbb R^{n \times n}$ is a square matrix:

  1. Either $Ax = b$ has a unique solution for all $b \in \Bbb R^{n \times 1}$, or
  2. For every $b \in \Bbb R^{n \times 1}$, $Ax = b$ either has no solution or infinite solutions.
$\endgroup$
2
  • $\begingroup$ in the given rref, the last line right hand side is not $c^2-c,$ correct is $c^2 - c - 2$ $\endgroup$
    – Will Jagy
    May 5 at 18:48
  • $\begingroup$ @WillJagy: Thank you, have changed it. (Again, I haven't checked it myself.) $\endgroup$ May 5 at 20:20

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.