1957 AMC 12 第 41 题

Question

给定方程组

\begin{aligned} ax+(a-1)y&=1,\\ (a+1)x-ay&=1 \end{aligned}\text{。}

当

a

取下列哪个值时，这个方程组没有关于

x

与

y

的解？

Given the system of equations

\begin{aligned} ax+(a-1)y&=1,\\ (a+1)x-ay&=1. \end{aligned}

For which one of the following values of

a

is there no solution for

x

and

y?

1

0

-1

\pm\dfrac{\sqrt2}{2}

\pm\sqrt2

当系数行列式为零时，一个

2\times2

方程组可能没有唯一解

A

2\times2

system can fail to have a unique solution when its coefficient determinant is zero

计算

a(-a)-(a-1)(a+1)

Compute

a(-a)-(a-1)(a+1)

系数行列式为

\begin{aligned} &a(-a)-(a-1)(a+1)\\ &\quad=1-2a^2 \end{aligned}\text{。}

当

a=\pm\frac{\sqrt2}{2}

时，它等于零。对这两个值中的任一个，两个系数行成比例，但右端的两个数不满足相同比例，所以方程组不相容。

因此，正确答案是 D。

The coefficient determinant is

\begin{aligned} &a(-a)-(a-1)(a+1)\\ &\quad=1-2a^2. \end{aligned}

It vanishes when

a=\pm\frac{\sqrt2}{2}.

For either value, the two coefficient rows are proportional but the two right sides are not in the same ratio, so the system is inconsistent.

Thus, the correct answer is D.

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