题目
11. (10分)设A=(}2&0&00&3&10&4&1,求X.
11. (10分)设$A=\left(\begin{matrix}2&0&0\\0&3&1\\0&4&1\end{matrix}\right),B=\left(\begin{matrix}1&0&0\\0&2&2\\0&2&1\end{matrix}\right),$并且$A^{*}$是A的伴随矩阵，矩阵X满足$AX+4E=BX+A^{*},$求X.
题目解答
答案
将方程整理为 $(A - B)X = A^* - 4E$。
计算得：
\[ A - B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 2 & 0 \end{pmatrix}, \quad (A - B)^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \\ 0 & -1 & \frac{1}{2} \end{pmatrix} \]
\[ A^* = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 2 & -2 \\ 0 & -8 & 6 \end{pmatrix}, \quad A^* - 4E = \begin{pmatrix} -5 & 0 & 0 \\ 0 & -2 & -2 \\ 0 & -8 & 2 \end{pmatrix} \]
\[ X = (A - B)^{-1}(A^* - 4E) = \boxed{\begin{pmatrix} -5 & -4 & -4 \\ -2 & -6 & 1 \\ 2 & -4 & 7 \end{pmatrix}} \]