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编写小组/试卷/1231线性代数考试卷（解析版）.md

@@ -125,12 +125,20 @@ $\quad T^{-1} = \begin{bmatrix} 1 & 0 \\ -5 & 1 \end{bmatrix}$是坐标变换矩
    $$
    \langle \alpha_1 + 2\alpha_2,\, 2\alpha_1 + \alpha_3 \rangle = \underline{\qquad\qquad}.
    $$
+---
+
+【答】4 
+
+【解析】由内积的性质可知 $$ \begin{aligned} \langle \alpha_1 + 2\alpha_2, 2\alpha_1 + \alpha_3 \rangle &= \langle \alpha_1, 2\alpha_1 + \alpha_3 \rangle + \langle 2\alpha_2, 2\alpha_1 + \alpha_3 \rangle \\[1em] &= \langle \alpha_1, 2\alpha_1 \rangle + \langle \alpha_1, \alpha_3 \rangle + \langle 2\alpha_2, 2\alpha_1 \rangle + \langle 2\alpha_2, \alpha_3 \rangle \\[1em] &= 2\langle \alpha_1, \alpha_1 \rangle + \langle \alpha_1, \alpha_3 \rangle + 4\langle \alpha_2, \alpha_1 \rangle + 2\langle \alpha_2, \alpha_3 \rangle, \end{aligned} $$ 再由题意，可知 $$ \langle \alpha_1, \alpha_1 \rangle = 2,\quad \langle \alpha_1, \alpha_3 \rangle = -2,\quad \langle \alpha_2, \alpha_1 \rangle = 2,\quad \langle \alpha_2, \alpha_3 \rangle = -3. $$ 从而 $$ \langle \alpha_1 + 2\alpha_2, 2\alpha_1 + \alpha_3 \rangle = 4. $$
+
+---
 
 8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\quad\quad}$。
 
 ---
 
 解析：
+
 先计算$A^2$：
 
 $$A^2