Is it sufficient?

Before Beginning we notice that the null matrix and the identity matrix(Also hinted at by part B) satisfy the definition in part A.

$A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

then $A^2=\begin{bmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{bmatrix}$

Since $A^2=A$

$\begin{bmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{bmatrix}=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

$\begin{cases} a^2+bc=a \\ b(a+d)=b \\ c(a+d)=c \\ d^2+bc=d\end{cases}$

Note that if $b$ or $c$ is 0 we have the null matrix.

Hence all we have to do is look for solutions when $a+d=1$

adding the $1^\text{st}$ and $4^\text{th}$ equation we have

$a^2+d^2+2bc=a+d$

$(a+d)^2-2da+2bc=(a+d)$

$ad=bc$

Hence we have the two conditions which will make $A^2=A$ , $a+d=1$ and $ad=bc$

Note that if $d=c=b=a=\frac{1}{2}$ we have $A^2=A$ , through the above condition. Hence $\mathbf B$ cannot be defined solely by condition $\mathbf A$ .