報載[1]：「大考中心審題老師說，相較數乙，數甲則是『平易近人』，近3年來最簡單，沒有複雜繁瑣的計算，只要題目讀過去，頭腦清楚就可以作答，預估五標都可能上升。」下面抄錄今年大學指考數學甲的一則線性代數問題 (2015 年大學指考數甲試題與解答)。
問題：考慮坐標平面上的直線 
![\left[\!\!\begin{array}{cr} 1&0\\ a&-8 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bcr%7D++1%260%5C%5C++a%26-8++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0&c=20201002)
(1) 6
(2) 8
(3) 10
(4) 12
(5) 14
解答：兩點決定一線。按照問題的陳述，找出直線 
![\displaystyle \left[\!\!\begin{array}{cr} 1&0\\ a&-8 \end{array}\!\!\right]\begin{bmatrix} 2\\ 3 \end{bmatrix}=\begin{bmatrix} 2\\ 2a-24 \end{bmatrix}=\alpha\begin{bmatrix} 1\\ 2 \end{bmatrix}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bcr%7D++1%260%5C%5C++a%26-8++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5Cbegin%7Bbmatrix%7D++2%5C%5C++3++%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D++2%5C%5C++2a-24++%5Cend%7Bbmatrix%7D%3D%5Calpha%5Cbegin%7Bbmatrix%7D++1%5C%5C++2++%5Cend%7Bbmatrix%7D&bg=ffffff&fg=000000&s=0&c=20201002)
解出 
具創意與想像力的學生可能採取這個跳躍解法：直線 ![\left[\!\!\begin{array}{r} 3\\ -2 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Br%7D++3%5C%5C++-2++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0&c=20201002)
![\left[\!\!\begin{array}{r} 2\\ -1 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Br%7D++2%5C%5C++-1++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0&c=20201002)
![\displaystyle \left[\!\!\begin{array}{cr} 1&0\\ a&-8 \end{array}\!\!\right]\left[\!\!\begin{array}{r} 3\\ -2 \end{array}\!\!\right]=\begin{bmatrix} 3\\ 3a+16 \end{bmatrix}=\beta\left[\!\!\begin{array}{r} 2\\ -1 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Bcr%7D++1%260%5C%5C++a%26-8++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Br%7D++3%5C%5C++-2++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%3D%5Cbegin%7Bbmatrix%7D++3%5C%5C++3a%2B16++%5Cend%7Bbmatrix%7D%3D%5Cbeta%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Br%7D++2%5C%5C++-1++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0&c=20201002)
解出 
![\displaystyle \left[\!\!\begin{array}{rr} 1&0\\ 14&-8 \end{array}\!\!\right]\left[\!\!\begin{array}{r} 3\\ -2 \end{array}\!\!\right]=\left[\!\!\begin{array}{r} 3\\ 58 \end{array}\!\!\right]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Brr%7D++1%260%5C%5C++14%26-8++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Br%7D++3%5C%5C++-2++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D%3D%5Cleft%5B%5C%21%5C%21%5Cbegin%7Barray%7D%7Br%7D++3%5C%5C++58++%5Cend%7Barray%7D%5C%21%5C%21%5Cright%5D&bg=ffffff&fg=000000&s=0&c=20201002)
這個現象的根本原因在於線性變換不總具有保角性──兩向量的夾角不因線性變換而改變。考試時，創意還是要服膺基本的數學規律。
設想，如果今年入闈出題的老師不是那麼「平易近人」，我們看到的考題可能如下：
問題：考慮坐標平面上的直線
(1) 6
(2) 8
(3) 10
(4) -35/6
(5) 14
我估計很多人都會猜
引用來源：
[1] 聯合新聞網：數學甲有鑑別度「比數乙簡單太多」。插一句話，太簡單的考題怎麼可能有鑑別度呢？
線代啟示錄 