system of equation (3 unknown, 3 equations)

Question

I want to solve a system of equations, but I seem to get it wrong.

Problem: see picture, and note that I should tell for which a I have

• no solution
• one solution
• $\infty$ solutions

Attempt:

$$\begin{matrix}i \\ ii \\ iii\end{matrix}\left\{\begin{matrix}x+y-az=3\\ ax-y+z=-2\\ -3x+y-z=-a+2\end{matrix}\right.$$

$ii+iii\to (a-3)x=-a\implies x=-\frac a{a-3}=\frac a{3-a}$

$i+ii\to (a+1)x+z(1-a)=1$ and $x=\frac a{3-a}\implies (a+1)\frac a{3-a}+z(1-a)=1\implies$

$$z=\frac {1-(a+1)\frac a{3-a}}{1-a}=\frac {a+3}{3-a}, a\ne 1,3$$

$$y=3-x+az=3-\frac a{3-a}+a\frac {a+3}{3-a}=\frac {3(3-a)-a(a+4)}{3-a},a\ne 1,3$$

Answer 1

Using Gaussian Elimination (your way works too) and reducing the augmented system, we get:

$$\begin{bmatrix}1 & 0 & 0 & -\dfrac{a}{a-3}\\0 & 1 & 0 & \dfrac{-a^2+a-9}{a-3} \\ 0 & 0 & 1 & \dfrac{a+3}{-a + 3}\end{bmatrix}$$

We found that $a = 1$ and $a = 3$ appeared to be problematic.

If we dig deeper, we actually find:

• $a = 3$ is no solution, because we get a RREF of:

$$\begin{bmatrix}1 & 0 & -\dfrac{1}{2} & 0\\0 & 1 & -\dfrac{5}{2} & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$$

• $a = 1$ is infinite solutions because we get a RREF of:

$$\begin{bmatrix}1 & 0 & 0 & \dfrac{1}{2}\\0 & 1 & -1 & \dfrac{5}{2} \\ 0 & 0 & 0 & 0\end{bmatrix}$$

This gives us: $x = \dfrac{1}{2}, y = \dfrac{5}{2} + z$, where $z$ is a free variable (hence, infinite solutions).

• All other $a$ (from the RREF above) give us a unique (single solution).