Example 10 Given $\frac{a_{2}+u_{3}+a_{4}}{a_{1}}=\frac{a_{1}+a_{3}+a_{4}}{a_{2}}$ $=\frac{a_{1}+a_{2}+a_{4}}{a_{3}}=\frac{a_{1}+a_{2}+a_{3}}{a_{4}}=k$. Then the value of $k$ is $(\quad)$.
(1995, Jiangsu Province Junior High School Mathematics Competition)
(A) 3
(B) $\frac{1}{3}$
(C) -1
(D) 3 or -1
Given the equations with five letters and four independent equations, it can be seen as a system of equations about $a_{1}, a_{2}, a_{3}, a_{4}$:
$$
\left\{\begin{array}{l}
a_{2}+a_{3}+a_{4}=a_{1} k, \\
a_{1}+a_{3}+a_{4}=a_{2} k, \\
a_{1}+a_{2}+a_{4}=a_{3} k, \\
a_{1}+a_{2}+a_{3}=a_{4} k .
\end{array}\right.
$$
(1) + (2) + (3) + (4) gives
$$
\begin{array}{l}
3\left(a_{1}+a_{2}+a_{3}+a_{4}\right) \\
=k\left(a_{1}+a_{2}+a_{3}+a_{4}\right) .
\end{array}
$$

If $a_{1}+a_{2}+a_{3}+a_{4} \neq 0$, then $k=3$;
If $a_{1}+a_{2}+a_{3}+a_{4}=0$, then $a_{1}=$ $-\left(a_{2}+a_{3}+a_{4}\right)$, and thus $k=-1$.
Therefore, $k=3$ or -1. Hence, the answer is (D).