I'm working through my linear algebra book and I'm having trouble with one of the questions. After searching for a bit on Google I found that someone else was having trouble with the same part and actually asked a question about it here. However, I don't think the answer is very clear (but I did use his method to try and solve the question), so I would like to confirm this is the correct way to do it.
Find all possible values of $rank(A)$ as $a$ varies:
$$ A= \begin{bmatrix} 1 & 2 & a \\ -2 & 4a & 2 \\ a & -2 & 1 \\ \end{bmatrix}, $$
$$ det(A) = -4a^3+12a+8 $$ Which gives $$ \begin{cases} a = -1\\ a = 2 \end{cases} $$ Substituting these in matrix $A$ and row reducing gives the two matrices
$$ (a=-1) = \begin{bmatrix} 1 & 2 & -1 \\[0.3em] 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 \end{bmatrix} ,% (a=2) = \begin{bmatrix} 1 & 0 & 1 \\[0.3em] 0 & 1 & \frac{1}{2} \\[0.3em] 0 & 0 & 0 \end{bmatrix} $$
And so if $a=-1$, then $rank(A) = 1$ and if $a=2$, then $rank(A)=2$. I assume this is the correct way to do this? The book also includes this is the answer: for $a \neq -1,2,$ $rank(A)=3$. Is this found by simply substituting $a$ with an arbitrary number that $\neq -1,2$, so for example $a=1$?
Note that if $a\neq -1,2$ $det(A)\neq 0$ therefore the matrix is invertible and the rank is $3$. Indeed $$ A= \begin{bmatrix} 1 & 2 & a \\ -2 & 4a & 2 \\ a & -2 & 1 \\ \end{bmatrix}, $$
$$ det(A) = -4a^3+12a+8 $$ If $a=-1$ the rank is $1$, if $a=2$ the rank is $2$