Finding out the number solutions for an equation system.

Question

I'm currently doing an exercise where I should solve the equation system:

ax + y + az = 2
x + ay + z = 2
x + az = 1

and then answer "for which values on a does the equation system only have on solution? For which values on a do we have an infinite amount of solutions, and what value on a lacks solutions.

After solving the equation system I got this as an answer:

Now, how do I solve for which values on a does the equation system only have on solution? For which values on a do we have an infinite amount of solutions, and what value on a lacks solutions.

Answer 1

Hint: From the last equation we get $$x=1-az$$ so we get

$$1-az+ay+z=2$$ or $$z(1-a)+ay=1$$ and with $$a(1-az)+y+az=2$$ we get $$y=2-a+a^2z-az$$ we get

$$z(1-a)+a(2-a+a^2z-a^2z)=2$$ Can you finish?

$$z(1-a+a^3-a^2)=2+a^2-2a$$ The last equation is given by $$z(1-a)^2(1+a)=a^2-2a+2$$

Answer 2

The determinant is given by

$$(a-1)^2 (a+1)$$

• When the $\det \ne 0$, we have a unique solution for a fixed $a$.

$$x = \dfrac{1}{a+1},~~ y = \frac{2}{a+1},~~ z = \dfrac{1}{a+1} $$

• When $a = -1$, the RREF yields no solution.

• When $a = 1$, the RREF yields infinite solutions and we have ($z$ is a free variable)

$$\begin{align} y &= 1 \\ x &= 1-z\end{align}$$