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111學測數學B試題-05


<單選>設矩陣 \( A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \),若 \( A^7 – 3A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \),則 \( a + b + c + d \) 之值為下列哪一個選項?
(1) -8
(2) -5
(3) 5
(4) 8
(5) 10

答案

\[
A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}
\]
那麼我們重新計算。

---

**1. 計算 \(A^2\)**
\[
A^2 = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}
= \begin{bmatrix} 1+1 & 1-1 \\ 1-1 & 1+1 \end{bmatrix}
= \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}
= 2I
\]

---

**2. 利用 \(A^2 = 2I\) 化簡 \(A^7\)**
\[
A^2 = 2I
\]
\[
A^4 = (A^2)^2 = (2I)^2 = 4I
\]
\[
A^6 = A^4 A^2 = (4I)(2I) = 8I
\]
\[
A^7 = A^6 \cdot A = (8I) A = 8A
\]

---

**3. 計算 \(A^7 - 3A\)**
\[
A^7 - 3A = 8A - 3A = 5A
\]
\[
5A = 5 \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}
= \begin{bmatrix} 5 & 5 \\ 5 & -5 \end{bmatrix}
\]

---

**4. 求 \(a+b+c+d\)**
\[
a=5, \quad b=5, \quad c=5, \quad d=-5
\]
\[
a+b+c+d = 5+5+5+(-5) = 10
\]

---

**答案:** (5) 10

---

所以正確的矩陣應是 \(A = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\),才會得到選項中的 10。 報錯
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