The linear system given is as follows:
ax +
y = 0
ay + z = 0
x + az = 0
I have attempted to solve this question. However, I am not really sure if my answer is correct as the question asks for "values" whereas after solving it I only got one value for a which would give me infinitely many solutions.
The value that I got was a = -1 as this was the only value that made the last row 0 = 0. Is this correct?
Best Answer
Since the question asks for values, our task well be to find the set of all $a$s such that this system has infinitely many solutions.
First, let's rewrite our system as $A_a \ X = 0$ where $X = (x, y, z)$ and $$A_a := \begin{pmatrix} a & 1 & 0 \\ 0 & a & 1 \\ 1 & 0 & a \end{pmatrix} $$ This system has infinitely many solutions iff $A_a$ isn't invertible, ie $\det(A_a)= 0$.
But we have $\det(A_a) = a^3 + 1$ : $A_a$ isn't invertible $\Longleftrightarrow$ $a$ is cubic root of $-1$.
Hence there is only one possible real value for $a$, namely $-1$.
There are two additional complex values for $a$, namely $e^{\frac{2 i \pi}{3}}$ and $e^{\frac{4 i \pi}{3}}$.