线代真题分块矩阵求伴随矩阵题目设A,BA,BA,B均为2阶矩阵A∗,B∗A^*,B^*A∗,B∗分别为A,BA,BA,B的伴随矩阵若∣A∣2,∣B∣3|A|2,|B|3∣A∣2,∣B∣3则分块矩阵M(0AB0)M\begin{pmatrix} 0 A \\ B 0 \end{pmatrix}M(0BA0)的伴随矩阵为(A)(03B∗2A∗0)\begin{pmatrix}0 3B^*\\2A^* 0\end{pmatrix}(02A∗3B∗0)(B)(02B∗3A∗0)\begin{pmatrix}0 2B^*\\3A^* 0\end{pmatrix}(03A∗2B∗0)©(02B3A0)\begin{pmatrix}0 2B\\3A 0\end{pmatrix}(03A2B0)(D)(03B2A0)\begin{pmatrix}0 3B\\2A 0\end{pmatrix}(02A3B0)答案B核心公式结论1伴随矩阵定义公式M∗∣M∣⋅M−1\boldsymbol{M^*|M|\cdot M^{-1}}M∗∣M∣⋅M−1结论2分块对角矩阵求逆(0AB0)−1(0B−1A−10)\begin{pmatrix}0 A\\B 0\end{pmatrix}^{-1} \begin{pmatrix}0 B^{-1}\\A^{-1} 0\end{pmatrix}(0BA0)−1(0A−1B−10)结论3方阵求逆与伴随关系A−11∣A∣A∗ ⟹ A∗∣A∣A−1\boldsymbol{A^{-1}\dfrac{1}{|A|}A^* \implies A^*|A|A^{-1}}A−1∣A∣1A∗⟹A∗∣A∣A−1分步计算步骤1计算分块矩阵MMM的行列式A,BA,BA,B都是2阶方阵∣M∣∣0AB0∣(−1)2×2∣A∣∣B∣(−1)4∣A∣∣B∣∣A∣∣B∣2×36 |M|\begin{vmatrix}0A\\B0\end{vmatrix}(-1)^{2\times2}|A||B|(-1)^4|A||B||A||B|2\times36∣M∣0BA0(−1)2×2∣A∣∣B∣(−1)4∣A∣∣B∣∣A∣∣B∣2×36分块交换行列式符号规则∣OABO∣(−1)mn∣A∣∣B∣\begin{vmatrix}OA\\BO\end{vmatrix}(-1)^{mn}|A||B|OBAO(−1)mn∣A∣∣B∣m,nm,nm,n是A、B的阶数这里m2,n2m2,n2m2,n2步骤2求MMM的逆矩阵$$M^{-1}\begin{pmatrix}0 A \B 0\end{pmatrix}^{-1}\begin{pmatrix}0 B^{-1} \A^{-1} 0\end{pmatrix}$$步骤3代入伴随公式M∗∣M∣M−1M^*|M|M^{-1}M∗∣M∣M−1$$M^* 6\cdot\begin{pmatrix}0 B^{-1} \A^{-1} 0\end{pmatrix}\begin{pmatrix}0 6B^{-1}\6A^{-1} 0\end{pmatrix}$$步骤4用A∗∣A∣A−1A^*|A|A^{-1}A∗∣A∣A−1替换逆矩阵A∗∣A∣A−12A−1 ⟹ A−112A∗A^*|A|A^{-1}2A^{-1} \implies \boldsymbol{A^{-1}\dfrac12 A^*}A∗∣A∣A−12A−1⟹A−121A∗B∗∣B∣B−13B−1 ⟹ B−113B∗B^*|B|B^{-1}3B^{-1} \implies \boldsymbol{B^{-1}\dfrac13 B^*}B∗∣B∣B−13B−1⟹B−131B∗代入6B−16⋅13B∗2B∗ 6B^{-1}6\cdot \frac13 B^*2B^*6B−16⋅31B∗2B∗6A−16⋅12A∗3A∗ 6A^{-1}6\cdot \frac12 A^*3A^*6A−16⋅21A∗3A∗得到M∗(02B∗3A∗0) M^*\begin{pmatrix} 0 2B^*\\ 3A^* 0 \end{pmatrix}M∗(03A∗2B∗0)对应选项B。易错点总结分块矩阵(OABO)\begin{pmatrix}OA\\BO\end{pmatrix}(OBAO)求逆不是直接(O,A−1B−1,O)\begin{pmatrix}O,A^{-1}\\B^{-1},O\end{pmatrix}(O,A−1B−1,O)A、B位置互换行列式符号(−1)mn(-1)^{mn}(−1)mnm,nm,nm,n是两个子矩阵的阶数不要忘记这个(−1)(-1)(−1)幂次。A∗A^*A∗和A−1A^{-1}A−1的转换A∗∣A∣A−1A^*|A|A^{-1}A∗∣A∣A−1是这道题的核心桥梁。速记结论(OABO)∗(O∣A∣B∗∣B∣A∗O) \boldsymbol{\begin{pmatrix}OA\\BO\end{pmatrix}^* \begin{pmatrix}O |A|B^*\\|B|A^* O\end{pmatrix}}(OBAO)∗(O∣B∣A∗∣A∣B∗O) 
长河君 · 2026/10/29 2:01:36 · 阅读 1.2k 线代真题｜分块矩阵求伴随矩阵
线代真题分块矩阵求伴随矩阵题目设A,BA,BA,B均为2阶矩阵A∗,B∗A^*,B^*A∗,B∗分别为A,BA,BA,B的伴随矩阵若∣A∣2,∣B∣3|A|2,|B|3∣A∣2,∣B∣3则分块矩阵M(0AB0)M\begin{pmatrix} 0 A \\ B 0 \end{pmatrix}M(0BA0)的伴随矩阵为(A)(03B∗2A∗0)\begin{pmatrix}0 3B^*\\2A^* 0\end{pmatrix}(02A∗3B∗0)(B)(02B∗3A∗0)\begin{pmatrix}0 2B^*\\3A^* 0\end{pmatrix}(03A∗2B∗0)©(02B3A0)\begin{pmatrix}0 2B\\3A 0\end{pmatrix}(03A2B0)(D)(03B2A0)\begin{pmatrix}0 3B\\2A 0\end{pmatrix}(02A3B0)答案B核心公式结论1伴随矩阵定义公式M∗∣M∣⋅M−1\boldsymbol{M^*|M|\cdot M^{-1}}M∗∣M∣⋅M−1结论2分块对角矩阵求逆(0AB0)−1(0B−1A−10)\begin{pmatrix}0 A\\B 0\end{pmatrix}^{-1} \begin{pmatrix}0 B^{-1}\\A^{-1} 0\end{pmatrix}(0BA0)−1(0A−1B−10)结论3方阵求逆与伴随关系A−11∣A∣A∗ ⟹ A∗∣A∣A−1\boldsymbol{A^{-1}\dfrac{1}{|A|}A^* \implies A^*|A|A^{-1}}A−1∣A∣1A∗⟹A∗∣A∣A−1分步计算步骤1计算分块矩阵MMM的行列式A,BA,BA,B都是2阶方阵∣M∣∣0AB0∣(−1)2×2∣A∣∣B∣(−1)4∣A∣∣B∣∣A∣∣B∣2×36 |M|\begin{vmatrix}0A\\B0\end{vmatrix}(-1)^{2\times2}|A||B|(-1)^4|A||B||A||B|2\times36∣M∣0BA0(−1)2×2∣A∣∣B∣(−1)4∣A∣∣B∣∣A∣∣B∣2×36分块交换行列式符号规则∣OABO∣(−1)mn∣A∣∣B∣\begin{vmatrix}OA\\BO\end{vmatrix}(-1)^{mn}|A||B|OBAO(−1)mn∣A∣∣B∣m,nm,nm,n是A、B的阶数这里m2,n2m2,n2m2,n2步骤2求MMM的逆矩阵$$M^{-1}\begin{pmatrix}0 A \B 0\end{pmatrix}^{-1}\begin{pmatrix}0 B^{-1} \A^{-1} 0\end{pmatrix}$$步骤3代入伴随公式M∗∣M∣M−1M^*|M|M^{-1}M∗∣M∣M−1$$M^* 6\cdot\begin{pmatrix}0 B^{-1} \A^{-1} 0\end{pmatrix}\begin{pmatrix}0 6B^{-1}\6A^{-1} 0\end{pmatrix}$$步骤4用A∗∣A∣A−1A^*|A|A^{-1}A∗∣A∣A−1替换逆矩阵A∗∣A∣A−12A−1 ⟹ A−112A∗A^*|A|A^{-1}2A^{-1} \implies \boldsymbol{A^{-1}\dfrac12 A^*}A∗∣A∣A−12A−1⟹A−121A∗B∗∣B∣B−13B−1 ⟹ B−113B∗B^*|B|B^{-1}3B^{-1} \implies \boldsymbol{B^{-1}\dfrac13 B^*}B∗∣B∣B−13B−1⟹B−131B∗代入6B−16⋅13B∗2B∗ 6B^{-1}6\cdot \frac13 B^*2B^*6B−16⋅31B∗2B∗6A−16⋅12A∗3A∗ 6A^{-1}6\cdot \frac12 A^*3A^*6A−16⋅21A∗3A∗得到M∗(02B∗3A∗0) M^*\begin{pmatrix} 0 2B^*\\ 3A^* 0 \end{pmatrix}M∗(03A∗2B∗0)对应选项B。易错点总结分块矩阵(OABO)\begin{pmatrix}OA\\BO\end{pmatrix}(OBAO)求逆不是直接(O,A−1B−1,O)\begin{pmatrix}O,A^{-1}\\B^{-1},O\end{pmatrix}(O,A−1B−1,O)A、B位置互换行列式符号(−1)mn(-1)^{mn}(−1)mnm,nm,nm,n是两个子矩阵的阶数不要忘记这个(−1)(-1)(−1)幂次。A∗A^*A∗和A−1A^{-1}A−1的转换A∗∣A∣A−1A^*|A|A^{-1}A∗∣A∣A−1是这道题的核心桥梁。速记结论(OABO)∗(O∣A∣B∗∣B∣A∗O) \boldsymbol{\begin{pmatrix}OA\\BO\end{pmatrix}^* \begin{pmatrix}O |A|B^*\\|B|A^* O\end{pmatrix}}(OBAO)∗(O∣B∣A∗∣A∣B∗O)