题号:6636    题型:解答题    来源:B站刘老师开讲《线性代数B》第七套期末模拟考试
已知 $\boldsymbol{A}, \boldsymbol{B}$ 均为三阶矩阵, $\boldsymbol{A}=\left(\begin{array}{rrr}1 & 1 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & -1\end{array}\right)$, 且满足 $\boldsymbol{A}^2-\boldsymbol{A} \boldsymbol{B}=\boldsymbol{E}$, 求矩阵 $\boldsymbol{B}$.
0 人点赞 纠错 ​ ​ 26 次查看 ​ 我来讲解
答案:
答案:
析 a. $A^2-A B=E \Rightarrow A(A-B)=E \Rightarrow A-B=A^{-1} \Rightarrow B=A-A^{-1}$;
b. $(\boldsymbol{A}, \boldsymbol{E}) \stackrel{r}{\sim}\left(E, A^{-1}\right)$.

解 因为 $A^2-A B=E$, (析 a) 所以 $B=A-A^{-1}$.
$$
\text { 由 }(\boldsymbol{A}, \boldsymbol{E})=\left(\begin{array}{rrr rrr}
1 & 1 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 & 1 & 0 \\
0 & 0 & -1 & 0 & 0 & 1
\end{array}\right) \text {, }
$$

$$
\left(\begin{array}{rrr rrr}
1 & 0 & -2 & 1 & -1 & 0 \\
0 & 1 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & -1
\end{array}\right)
$$

$$
\left(\begin{array}{rrrrrr}
1 & 0 & 0 & 1 & -1 & 2 \\
0 & 1 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 0 & -1
\end{array}\right)
$$


$$
A^{-1}=\left(\begin{array}{rrr}
1 & -1 & -2 \\
0 & 1 & 1 \\
0 & 0 & -1
\end{array}\right),
$$ 于是

$$
\boldsymbol{B}=\boldsymbol{A}-\boldsymbol{A}^{-1}=\left(\begin{array}{rrr}
1 & 1 & -1 \\
0 & 1 & 1 \\
0 & 0 & -1
\end{array}\right)-\left(\begin{array}{rrr}
1 & -1 & -2 \\
0 & 1 & 1 \\
0 & 0 & -1
\end{array}\right)=\left(\begin{array}{lll}
0 & 2 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right) .
$$

关闭页面 下载Word格式