## 2017 年大學學測的線性代數問題

$a_1,\ldots,a_9$ 為等差數列，且 $k$ 為實數，若方程組

\left\{\begin{aligned} a_1x-a_2y+2a_3z&=k+1\\ a_4x-a_5y+2a_6z&=-k-5\\ a_7x-a_8y+2a_9z&=k+9 \end{aligned}\right.

\begin{aligned} (a_1+a_7)x-(a_2+a_8)y+2(a_3+a_9)z=2k+10&\Rightarrow 2(a_4x-a_5y+2a_6z)=2(k+5)\\ &\equiv a_4x-a_5y+2a_6z=-k-5 \end{aligned}

\begin{aligned} &\left[\!\!\begin{array}{ccccr} a_1&-a_2&2a_3&\vline&k+1\\ a_4&-a_5&2a_6&\vline&-k-5\\ a_7&-a_8&2a_9&\vline&k+9 \end{array}\!\!\right]\\ &\to\left[\!\!\begin{array}{rrrcr} a_1&-a_2&2a_3&\vline&k+1\\ a_7&-a_8&2a_9&\vline&k+9\\ a_4&-a_5&2a_6&\vline&-k-5 \end{array}\!\!\right]\\ &\to\left[\!\!\begin{array}{rrrcr} a_1&-a_2&2a_3&\vline&k+1\\ a_7&-a_8&2a_9&\vline&k+9\\ 2a_4&-2a_5&4a_6&\vline&-2k-10 \end{array}\!\!\right]\\ &\to\left[\!\!\begin{array}{ccccc} a_1&-a_2&2a_3&\vline&k+1\\ a_7&-a_8&2a_9&\vline&k+9\\ 2a_4-a_1-a_7&-2a_5+a_2+a_8&4a_6-2a_3-2a_9&\vline&-2k-10-(k+1)-(k+9)\end{array}\!\!\right]\\ &\to\left[\!\!\begin{array}{ccccc} a_1&-a_2&2a_3&\vline&k+1\\ a_7&-a_8&2a_9&\vline&k+9\\ 0&0&0&\vline&-4(k+5)\end{array}\!\!\right]\end{aligned}

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### 2 Responses to 2017 年大學學測的線性代數問題

1. 周伯欣 says:

謝謝周老師指教！

• Meiyue Shao says:

从逻辑完整性的角度讲还需要再验证 k=-5 时原方程确实有解，而不能满足于一个未经过检验的必要条件。