若實數 \(a, b, c, d\) 使得聯立方程組 \(\begin{cases} ax + 8y = c \\x – 4y = 3 \end{cases}\) 有解,且聯立方程組 \(\begin{cases} -3x + by = d\\ x – 4y = 3 \end{cases}\) 無解,則下列哪些選項一定正確？
(1) \(a \neq -2\)
(2) \(c = -6\)
(3) \(b = 12\)
(4) \(d \neq -9\)
(5) 聯立方程組 \(\begin{cases} ax + 8y = c \\ -3x + by = d \end{cases}\) 無解
\[
\boxed{\text{問題分析：二元一次聯立方程組解的判別}}
\]
\[
\begin{aligned}
& \text{考慮方程組 (I): }
\begin{cases}
ax + 8y = c \\
x - 4y = 3
\end{cases} \quad \text{有解} \\
\\
& \text{設 } L_1: ax + 8y = c,\quad L_2: x - 4y = 3 \\
\\
& \text{情況 ① 恰有一組解：} \\
& \quad \frac{a}{1} \neq \frac{8}{-4} \quad \Rightarrow \quad a \neq -2 \\
\\
& \text{情況 ② 無限多組解：} \\
& \quad \frac{a}{1} = \frac{8}{-4} = \frac{c}{3} \quad \Rightarrow \quad a = -2,\ c = -6 \\
\\
& \therefore \text{方程組 (I) 有解 } \Rightarrow a = -2 \text{ 或 } a \neq -2 \text{（但必定有解）}
\end{aligned}
\]
\[
\boxed{\text{第二組方程分析}}
\]
\[
\begin{aligned}
& \text{考慮方程組 (II): }
\begin{cases}
-3x + by = d \\
x - 4y = 3
\end{cases} \quad \text{無解} \\
\\
& \text{設 } L_3: -3x + by = d,\quad L_4: x - 4y = 3 \\
\\
& \text{無解的條件：} \\
& \quad \frac{-3}{1} = \frac{b}{-4} \neq \frac{d}{3} \\
\\
& \Rightarrow \frac{-3}{1} = \frac{b}{-4} \quad \Rightarrow \quad b = 12 \\
& \quad \text{且 } \frac{b}{-4} \neq \frac{d}{3} \quad \Rightarrow \quad \frac{12}{-4} \neq \frac{d}{3} \quad \Rightarrow \quad d \neq -9
\end{aligned}
\]
\[
\boxed{\text{綜合分析與判斷}}
\]
\[
\begin{aligned}
& \text{現在考慮聯立方程組：}
\begin{cases}
ax + 8y = c \\
-3x + 12y = d \quad (b=12)
\end{cases} \\
\\
& \text{係數比分析：} \\
& \quad \frac{a}{-3} \ \text{與} \ \frac{8}{12} = \frac{2}{3} \ \text{的關係決定解的情況} \\
\\
& \text{可能情況：} \\
& \bullet \ \text{若 } \frac{a}{-3} \neq \frac{2}{3} \ \Rightarrow \ \text{恰有一組解} \\
& \bullet \ \text{若 } \frac{a}{-3} = \frac{2}{3} \ \text{且} \ \frac{c}{d} \neq \frac{2}{3} \ \Rightarrow \ \text{無解} \\
& \bullet \ \text{若 } \frac{a}{-3} = \frac{2}{3} \ \text{且} \ \frac{c}{d} = \frac{2}{3} \ \Rightarrow \ \text{無限多組解} \\
\\
& \text{已知條件約束：} \\
& \quad a = -2 \ (\text{無限多解情況}) \ \text{或} \ a \neq -2 \\
& \quad b = 12,\ d \neq -9 \\
\\
& \text{代入 } a = -2: \ \frac{-2}{-3} = \frac{2}{3} = \frac{8}{12} \\
& \quad \text{此時若 } \frac{c}{d} = \frac{2}{3} \ \Rightarrow \ \text{無限多解（但需檢查是否可能）} \\
& \quad \text{若 } \frac{c}{d} \neq \frac{2}{3} \ \Rightarrow \ \text{無解} \\
\\
& \text{當 } a \neq -2 \ \text{時：} \ \frac{a}{-3} \neq \frac{2}{3} \ \Rightarrow \ \text{恰有一組解} \\
\\
& \therefore \text{最終方程組可能的情況為：} \\
& \quad \text{(3) 可能無解} \\
& \quad \text{(4) 可能恰有一組解}
\end{aligned}
\]
\[
\boxed{\text{結論：選項 (3) 和 (4) 正確}}
\]
\[
\begin{array}{c|c}
\text{條件} & \text{解的情況} \\ \hline
a = -2 \ \text{且} \ \dfrac{c}{d} = \dfrac{2}{3} & \text{無限多組解} \\
a = -2 \ \text{且} \ \dfrac{c}{d} \neq \dfrac{2}{3} & \text{無解} \\
a \neq -2 & \text{恰有一組解} \\
\end{array}
\]
\[
\boxed{\text{※ 關鍵要點：二元一次方程組解的判別}}
\]
\[
\begin{aligned}
\text{設方程組：} &
\begin{cases}
a_1x + b_1y = c_1 \\
a_2x + b_2y = c_2
\end{cases} \\
\\
\text{解的判別：} & \quad
\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \ \Rightarrow \ \text{恰有一組解} \\
& \quad
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \ \Rightarrow \ \text{無限多組解} \\
& \quad
\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \ \Rightarrow \ \text{無解}
\end{aligned}
\]
