$4 \cdot 242$ Let $a, b$ be integers, what is the smallest positive integer value $c$ of $\Delta=\left|\begin{array}{ll}36 & b \\ 81 & a\end{array}\right|$? For the positive integer solutions $(a, b)$ that satisfy $\Delta=c$, to make $a+b$ the smallest, what are $a, b$ respectively?
［Solution］ Since $\Delta=36 a-81 b=9(4 a-9 b)$,
and $a, b$ are integers, hence $4 a-9 b$ is also an integer. Since the smallest positive integer is 1, the smallest positive value $c$ of $\Delta$ is no less than 9.
Since 4 and 9 are coprime, by the theorem, there must exist integers $a, b$ such that
$$
\begin{array}{c}
4 a-9 b=1, \\
\text { by } a=\frac{9 b+1}{4}=2 b+\frac{b+1}{4},
\end{array}
$$

For $a$ to be an integer, $\frac{b+1}{4}$ must be an integer,
$$
\text { let } t=\frac{b+1}{4} \text {, }
$$

Thus, $\left\{\begin{array}{l}b=4 t-1, \\ a=2(4 t-1)+t=9 t-2 .(t=0, \pm 1, \pm 2, \cdots)\end{array}\right.$
When $t$ takes positive integers $1,2 \cdots$, all positive integer solutions are obtained.
$$
\text { and } a+b=(4 t-1)+(9 t-2)=13 t-3(t=1,2, \cdots)
$$

Therefore, taking $t=1$ makes $a+b$ the smallest value.
At this point, $a=7, b=3, a+b=10$.