Answer
See below.
Work Step by Step
$\begin{bmatrix}
1& 0 \\
a&b\\
\end{bmatrix} \begin{bmatrix}
1& 0 \\
a&b\\
\end{bmatrix}= \begin{bmatrix}
1\cdot1+0\cdot a& 1\cdot0+0\cdot b \\
a\cdot1+b\cdot a &0\cdot a+b\cdot b\\
\end{bmatrix}=\begin{bmatrix}
1& 0 \\
a+ab&b^2\\
\end{bmatrix}$
Thus we have two equations to satisfy: $a=a+ab$ and $b=b^2$.
Thus we have $2$ cases:
1) $b=1$, the $a=a+a(1)=a+a=2a$, thus $b=1,a=0$.
2) $a=a+a(0)=a$, thus $a$ is a free variable, so the solution here is: $b=0,a$.
