# Find the values of $c$ (if they exist) such that the following system has no, unique, and infinite solutions.

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

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 $$[0 0 0 | 0]$$)

• the last line right hand side is not $c^2-c,$ it is $c^2 - c - 2$ May 5 at 18:43
• 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 May 5 at 20:59

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.
• in the given rref, the last line right hand side is not $c^2-c,$ correct is $c^2 - c - 2$ May 5 at 18:48