kaoyan3basic 线性代数 第356题
📝 题目
### 第356题 356 （2018，数农）矩阵 $\left[\begin{array}{lll}0 & 0 & a \\ 0 & b & 0 \\ c & 0 & 0\end{array}\right]$ 的伴随矩阵为 （A）$\left[\begin{array}{ccc}0 & 0 & -b c \\ 0 & -a c & 0 \\ -a b & 0 & 0\end{array}\right]$ ． （B）$\left[\begin{array}{ccc}0 & 0 & -a b \\ 0 & -a c & 0 \\ -b c & 0 & 0\end{array}\right]$ ． （C）$\left[\begin{array}{ccc}0 & 0 & -b c \\ 0 & a c & 0 \\ -a b & 0 & 0\end{array}\right]$ ． （D）$\left[\begin{array}{ccc}0 & 0 & -a b \\ 0 & a c & 0 \\ -b c & 0 & 0\end{array}\right]$ ．
💡 答案解析
**答案**：A **解析**：步骤1：设原矩阵为$A$，计算行列式$|A| = -abc$。步骤2：计算各元素的代数余子式：$A_{11} = (-1)^{1+1} \begin{vmatrix} b & 0 \\ 0 & 0 \end{vmatrix} = 0$，$A_{12} = (-1)^{1+2} \begin{vmatrix} 0 & 0 \\ c & 0 \end{vmatrix} = 0$，$A_{13} = (-1)^{1+3} \begin{vmatrix} 0 & b \\ c & 0 \end{vmatrix} = -bc$；$A_{21} = (-1)^{2+1} \begin{vmatrix} 0 & a \\ 0 & 0 \end{vmatrix} = 0$，$A_{22} = (-1)^{2+2} \begin{vmatrix} 0 & a \\ c & 0 \end{vmatrix} = -ac$，$A_{23} = (-1)^{2+3} \begin{vmatrix} 0 & 0 \\ c & 0 \end{vmatrix} = 0$；$A_{31} = (-1)^{3+1} \begin{vmatrix} 0 & a \\ b & 0 \end{vmatrix} = -ab$，$A_{32} = (-1)^{3+2} \begin{vmatrix} 0 & a \\ 0 & 0 \end{vmatrix} = 0$，$A_{33} = (-1)^{3+3} \begin{vmatrix} 0 & 0 \\ 0 & b \end{vmatrix} = 0$。步骤3：伴随矩阵$A^* = (A_{ij})^T = \begin{bmatrix} 0 & 0 & -bc \\ 0 & -ac & 0 \\ -ab & 0 & 0 \end{bmatrix}$。 **难度**：★★☆☆☆