- if $a=-1$, then $\operatorname{rank}(A)=1$;
- if $a=2$, then $\operatorname{rank}(A)=2$;
- otherwise $\operatorname{rank}(A)=3$.
Using the elimination method, we obtain:
$$
A=\left[\begin{array}{ccc}
1 & 2 & a \\
-2 & 4 a & 2 \\
a & -2 & 1
\end{array}\right] \stackrel{R_{2}+2 R_{1}}{\stackrel{R_{3}-a R_{1}}{\longrightarrow}}\left[\begin{array}{ccc}
1 & 2 & a \\
0 & 4 a+4 & 2+2 a \\
0 & -2-2 a & 1-a^{2}
\end{array}\right]=B
$$
Let us consider two cases.
Case $1: a=-1$. Then the matrix $B$ is equal to
$$
\left[\begin{array}{lll}
1 & 2 & a \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] \text {. }
$$
Therefore, $B$ (and hence $A$ ) has rank $1.$
Case $2: a \neq-1$. Then we divide the second and the third rows of $B$ by $4 a+4$ and $-2-2 a$ respectively:
$$
\left[\begin{array}{ccc}
1 & 2 & a \\
0 & 4 a+4 & 2+2 a \\
0 & -2-2 a & 1-a^{2}
\end{array}\right] \stackrel{\stackrel{R_{2} /(4 a+4)}{R_{3} /(-2-a)}}{{\longrightarrow}}\left[\begin{array}{ccc}
1 & 2 & a \\
0 & 1 & 1 / 2 \\
0 & 1 & \frac{(1-a)(1+a)}{-2-2 a} \\
\end{array}\right]=\left[\begin{array}{ccc}
1 & 2 & a \\
0 & 1 & 1 / 2 \\
0 & 1 & \frac{4-1}{2}
\end{array}\right] \stackrel{R_{3}-R_{2}}{\longrightarrow} \left[\begin{array}{ccc}
1 & 2 & a \\
0 & 1 & 1 / 2 \\
0 & 0 & \frac{a-2}{2}
\end{array}\right]=C .
$$
Let us again consider two cases.
Case $2a: a=2$. Then
$$
\left[\begin{array}{ccc}
1 & 2 & a \\
0 & 1 & 1 / 2 \\
0 & 0 & \frac{a-2}{2}
\end{array}\right]=\left[\begin{array}{ccc}
1 & 2 & a \\
0 & 1 & 1 / 2 \\
0 & 0 & 0
\end{array}\right]
$$
has rank $2 .$
Case $2b: a \neq 2$. Then
$$
\left[\begin{array}{ccc}
1 & 2 & a \\
0 & 1 & 1 / 2 \\
0 & 0 & \frac{a-2}{2}
\end{array}\right] \stackrel{R_{3} / \frac{a-2}{2}}{\longrightarrow}\left[\begin{array}{ccc}
1 & 2 & a \\
0 & 1 & 1 / 2 \\
0 & 0 & 1
\end{array}\right]
$$
The last matrix has rank $3 .$
