Example 1 Solve the system of congruences
$$\left\{\begin{array}{l}
x \equiv 1(\bmod 3) \\
x \equiv-1(\bmod 5) \\
x \equiv 2(\bmod 7) \\
x \equiv-2(\bmod 11)
\end{array}\right.$$
Let $m_{1}=3, m_{2}=5, m_{3}=7, m_{4}=11$, then $M_{1}=5 \cdot 7 \cdot 11, M_{2}=3 \cdot 7 \cdot 11, M_{3}=$ $3 \cdot 5 \cdot 11, M_{4}=3 \cdot 5 \cdot 7$

From $M_{1} \equiv(-1) \cdot(1) \cdot(-1) \equiv 1(\bmod 3)$,
so $1 \equiv M_{1} M_{1}^{-1} \equiv M_{1}^{-1}(\bmod 3)$, we can take $M_{1}^{-1}=1$.
From $M_{2} \equiv(-2) \cdot(2) \cdot(1) \equiv 1(\bmod 5)$,
so $1 \equiv M_{2} M_{2}^{-1} \equiv M_{2}^{-1}(\bmod 5)$, we can take $M_{2}^{-1}=1$.
From $M_{3} \equiv 3 \times 5 \times 4 \equiv 4(\bmod 7)$,
so $1 \equiv M_{3} M_{3}^{-1} \equiv 4 M_{3}^{-1}(\bmod 7)$, we can take $M_{3}^{-1}=2$.
From $M_{4} \equiv 3 \times 5 \times 7 \equiv 4 \times 7 \equiv 6(\bmod 11)$,
so $1 \equiv M_{4} M_{4}^{-1} \equiv 6 M_{4}^{-1}(\bmod 11)$, we can take $M_{4}^{-1}=2$.
By the Chinese Remainder Theorem, the solution to the system of congruences is
$$x=(5 \cdot 7 \cdot 11) \cdot 1 \cdot 1+(3 \cdot 7 \cdot 11) \cdot 1 \cdot(-1)+(3 \cdot 5 \cdot 11) \cdot 2 \cdot 2+(3 \cdot 5 \cdot 7)$$
- $2 \cdot(-2)(\bmod 3 \cdot 5 \cdot 7 \cdot 11)$

which is $x \equiv 394(\bmod 1155)$.