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 $m_{1}=3, m_{2}=5, m_{3}=7, m_{4}=11$, satisfying the conditions of Theorem 1. In this case, $M_{1}=5 \cdot 7 \cdot 11, M_{2}=3 \cdot 7 \cdot 11, M_{3}=3 \cdot 5 \cdot 11, M_{4}=3 \cdot 5 \cdot 7$. We will find $M_{j}^{-1}$. Since $M_{1} \equiv(-1) \cdot(1) \cdot(-1) \equiv 1(\bmod 3)$, we have
$$1 \equiv M_{1} M_{1}^{-1} \equiv M_{1}^{-1}(\bmod 3)$$

Therefore, we can take $M_{1}^{-1}=1$. From $M_{2} \equiv(-2) \cdot(2) \cdot 1 \equiv 1(\bmod 5)$, we have
$$1 \equiv M_{2} M_{2}^{-1} \equiv M_{2}^{-1}(\bmod 5)$$

Therefore, we can take $M_{2}^{-1}=1$. From $M_{3} \equiv 3 \cdot 5 \cdot 4 \equiv 4(\bmod 7)$, we have
$$1 \equiv M_{3} M_{3}^{-1} \equiv 4 M_{3}^{-1}(\bmod 7)$$

Therefore, we can take $M_{3}^{-1}=2$. From $M_{4} \equiv 3 \cdot 5 \cdot 7 \equiv 4 \cdot 7 \equiv 6(\bmod 11)$, we have
$$1 \equiv M_{4} M_{4}^{-1} \equiv 6 M_{4}^{-1}(\bmod 11)$$

Therefore, we can take $M_{4}^{-1}=2$. Thus, by Theorem 1, the solution to the system of congruences is
$$\begin{aligned}
x \equiv & (5 \cdot 7 \cdot 11) \cdot 1 \cdot 1+(3 \cdot 7 \cdot 11) \cdot 1 \cdot(-1)+(3 \cdot 5 \cdot 11) \cdot 2 \cdot 2 \\
& +(3 \cdot 5 \cdot 7) \cdot 2 \cdot(-2)(\bmod 3 \cdot 5 \cdot 7 \cdot 11),
\end{aligned}$$

i.e., $\square$
$$x \equiv 385-231+660-420 \equiv 394(\bmod 1155)$$