设 $A, B$ 为 $n$ 阶可逆矩阵, $E$ 为 $n$ 阶单位矩阵, $M$ 为矩阵 $M$ 的伴随矩阵, 则 $\left(\begin{array}{ll}A & E \\ O & B\end{array}\right)^{\circ}= $
$ \text{A.} $ $\left(\begin{array}{cc}|A| B^* & -B^* A^{\circ} \\ 0 & A^* B^*\end{array}\right)$ $ \text{B.} $ $\left(\begin{array}{cc}|A| B^* & -A^* B^* \\ 0 & |B| A^*\end{array}\right)$ $ \text{C.} $ $\left(\begin{array}{cc}|B| A^* & -B^* A^* \\ 0 & |A| B^*\end{array}\right)$ $ \text{D.} $ $\left(\begin{array}{cc}|B| A^* & -A^* B^* \\ 0 & |A|^*\end{array}\right)$
【答案】 D

【解析】 结合伴随矩阵的核心公式, 代入(D)计算知
$$
\begin{aligned}
& \left(\begin{array}{cc}
A & E \\
O & B
\end{array}\right)\left(\begin{array}{cc}
|B| A^* & -A^* B^* \\
O & |A| B^*
\end{array}\right)=\left(\begin{array}{cc}
|B| A A^* & -A A^* B^*+|A| B^* \\
O & |A| B B^*
\end{array}\right) \\
& =\left(\begin{array}{cc}
|B||A| E & -|A| B^*+|A| B^* \\
O & |A||B| E
\end{array}\right)=\left(\begin{array}{cc}
|B||A| E & O \\
O & |A||B| E
\end{array}\right)=|A||B| E_{2 n} \text {, 故(D)正确. }
\end{aligned}
$$
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