$3 \cdot 2$ If $n$ is a real number, then the system of equations
$$
\left\{\begin{array}{l}
n x+y=1, \\
n y+z=1, \\
x+n z=1 .
\end{array}\right.
$$

has no solution if and only if the value of $n$ is
(A) -1 .
(B) 0 .
(C) 1 .
(D) 0 or 1 .
(E) $\frac{1}{2}$.
(24th American High School Mathematics Examination, 1973)
[Solution] If $n \neq -1$, using the method of addition, subtraction, or substitution, we can solve to get
$$
(x, y, z)=\left(\frac{1}{n+1}, \frac{1}{n+1}, \frac{1}{n+1}\right) \text {, }
$$

This is the solution to the known system of equations.
If $n=-1$, adding the two sides of the three equations yields $0=3$, a contradiction! Therefore, the answer is $(A)$.