1973 AMC 12 第 10 题

Question

若

n

为实数，则联立方程组

\begin{cases} nx+y=1,\\ ny+z=1,\\ x+nz=1 \end{cases}

无解当且仅当

n

等于

If

n

is a real number, then the simultaneous system

\begin{cases} nx+y=1,\\ ny+z=1,\\ x+nz=1 \end{cases}

has no solution if and only if

n

is equal to

-1

0

1

0

或

1

0

or

1

\frac12

将三个方程相加，观察

x+y+z

的系数

Add all three equations and inspect the coefficient of

x+y+z

对其余每个

n

值，尝试对称取值

x=y=z

For every remaining value of

n,

try the symmetric choice

x=y=z

将三个方程相加，得到

(n+1)(x+y+z)=3\text{。}

当

n=-1

时，这变为

0=3

，所以无解。若

n\ne-1

，对称取值

x=y=z=\frac1{n+1}

满足全部三个方程。因此方程组恰在

n=-1

时无解。

所以正确答案是 A。

Adding the equations gives

(n+1)(x+y+z)=3.

When

n=-1,

this becomes

0=3,

so no solution exists. If

n\ne-1,

the symmetric assignment

x=y=z=\frac1{n+1}

satisfies all three equations. Hence the system has no solution exactly when

n=-1.

Therefore, the correct answer is A.

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