Example 5 Given that $a, b, x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
a x+b y=3, \\
a x^{2}+b y^{2}=7, \\
a x^{3}+b y^{3}=16, \\
a x^{4}+b y^{4}=42 .
\end{array}\right.
$$

Find the value of $a x^{5}+b y^{5}$.
Given the conditions, construct the recurrence relation:
$$
\begin{array}{l}
s_{n}=a x^{n}+b y^{n}, n=1,2, \cdots, \\
\text { then } s_{n+2}=a x^{n+2}+b y^{n+2} \\
=\left(a x^{n+1}+b y^{n+1}\right)(x+y)- \\
\quad x y\left(a x^{n}+b y^{n}\right) .
\end{array}
$$

Therefore, \( s_{n+2}=(x+y) s_{n+1}-x y s_{n} \). Substituting the known equations, we get the system of equations:
$$
\left\{\begin{array}{l}
16=7(x+y)-3 x y, \\
42=16(x+y)-7 x y .
\end{array}\right.
$$

Solving these, we get \( x+y=-14 \) and \( x y=-38 \).
Thus, \( s_{5}=(x+y) s_{4}-x y s_{3} \)
$$
=-14 \times 42+38 \times 16=20 \text {, }
$$

which means \( a x^{5}+b y^{5}=20 \).