= Solution
Squaring with the product law gives
$$
A(a,b,c)^2=A(2a,\ 2b+ac,\ 2c).
$$
Equality to the <identity matrix> requires $2a=2c=0$ in $\mathbb F_5$. Since 2 has a multiplicative inverse modulo 5, this forces $a=c=0$. The middle condition then gives $2b=0$, hence $b=0$. Therefore
$$
\boxed{\{A\in G:A^2=I\}=\{I\}.}
$$
This agrees with <Lagrange's theorem>: a group of odd order has no element of order two.
The subset specified by $b\ne0$ \b[is not a subgroup]. It omits the <identity element>. It also fails <closure>: both $A(0,1,0)$ and $A(0,4,0)$ lie in it, but their product is $A(0,0,0)=I$, whose $b$ coordinate is zero.
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