網友周伯欣轉來2017年大學學測的一道線性代數問題：
設 
有解，則
網友周伯欣分享了他的解答：https://goo.gl/photos/WVfF3Kg5LzWNcHUSA並問道：周老師有興趣談談今年大學學測的一次方程組題目嗎？
既然這是大學學測的考題，我應該設想自己是一個正準備於今年秋季進入大學校園的有為青年 (但我要承認這有點困難，今早天氣稍微轉涼起床時差一點閃到腰)，白話文是我還沒學過大學線性代數 (當然也沒看過周老師的教學光碟)。閒話休提，第一句話說 
底下是我的解答：如果我們相信孟子的人性本善說，那麼出題老師必定能體諒應考學子的辛勞，不太可能出一道需要龐大計算的問題。因此，我假設出題老師是一位慈眉善目的長者，他先透露方程組可能無解，暗示我們至少有一條方程式是多餘的。不僅如此，這位慈善的長者還特意安排方程組的係數按順序排列 
也就是說 
註解
如果問題改成：求出 
![\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}](https://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D+%26%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bccccr%7D+a_1%26-a_2%262a_3%26%5Cvline%26k%2B1%5C%5C+a_4%26-a_5%262a_6%26%5Cvline%26-k-5%5C%5C+a_7%26-a_8%262a_9%26%5Cvline%26k%2B9+%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5C%5C+%26%5Cto%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brrrcr%7D+a_1%26-a_2%262a_3%26%5Cvline%26k%2B1%5C%5C+a_7%26-a_8%262a_9%26%5Cvline%26k%2B9%5C%5C+a_4%26-a_5%262a_6%26%5Cvline%26-k-5+%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5C%5C+%26%5Cto%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brrrcr%7D+a_1%26-a_2%262a_3%26%5Cvline%26k%2B1%5C%5C+a_7%26-a_8%262a_9%26%5Cvline%26k%2B9%5C%5C+2a_4%26-2a_5%264a_6%26%5Cvline%26-2k-10+%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5C%5C+%26%5Cto%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bccccc%7D+a_1%26-a_2%262a_3%26%5Cvline%26k%2B1%5C%5C+a_7%26-a_8%262a_9%26%5Cvline%26k%2B9%5C%5C+2a_4-a_1-a_7%26-2a_5%2Ba_2%2Ba_8%264a_6-2a_3-2a_9%26%5Cvline%26-2k-10-%28k%2B1%29-%28k%2B9%29%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5C%5C+%26%5Cto%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bccccc%7D+a_1%26-a_2%262a_3%26%5Cvline%26k%2B1%5C%5C+a_7%26-a_8%262a_9%26%5Cvline%26k%2B9%5C%5C+0%260%260%26%5Cvline%26-4%28k%2B5%29%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5Cend%7Baligned%7D&bg=ffffff&fg=000000&s=0&c=20201002)
上面最後一個步驟使用了等差數列性質 
線代啟示錄 
謝謝周老師指教！
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。