题号:
6636
题型:
解答题
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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 $\boldsymbol{B}$.
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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 $B=A-A^{-1}$.
$$
\text { 由 }(\boldsymbol{A}, \boldsymbol{E})=\left(\begin{array}{rrr rrr}
1 & 1 & -1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 & 1 & 0 \\
0 & 0 & -1 & 0 & 0 & 1
\end{array}\right) \text {, }
$$
$$
\left(\begin{array}{rrr rrr}
1 & 0 & -2 & 1 & -1 & 0 \\
0 & 1 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & -1
\end{array}\right)
$$
$$
\left(\begin{array}{rrrrrr}
1 & 0 & 0 & 1 & -1 & 2 \\
0 & 1 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 0 & -1
\end{array}\right)
$$
得
$$
A^{-1}=\left(\begin{array}{rrr}
1 & -1 & -2 \\
0 & 1 & 1 \\
0 & 0 & -1
\end{array}\right),
$$ 于是
$$
\boldsymbol{B}=\boldsymbol{A}-\boldsymbol{A}^{-1}=\left(\begin{array}{rrr}
1 & 1 & -1 \\
0 & 1 & 1 \\
0 & 0 & -1
\end{array}\right)-\left(\begin{array}{rrr}
1 & -1 & -2 \\
0 & 1 & 1 \\
0 & 0 & -1
\end{array}\right)=\left(\begin{array}{lll}
0 & 2 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right) .
$$
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