设 $f(x)=3-7 x-x^{2}, \quad A=\left(\begin{array}{lll}-3 & 4 & 0 \\ 1 & -5 & 1 \\ 2 & 0 & -7\end{array}\right)$, 求 $f(A)$
解： 由 $A^{2}=\left(\begin{array}{ccc}13 & -32 & 4 \\ -6 & 29 & -12 \\ -20 & 8 & 49\end{array}\right)$
得 $f(A)=3 E-7 A-A^{2}=$ $\left(\begin{array}{lll}3 & & \\ & 3 & \\ & & 3\end{array}\right)-7\left(\begin{array}{ccc}-3 & 4 & 0 \\ 1 & -5 & 1 \\ 2 & 0 & -7\end{array}\right)-\left(\begin{array}{ccc}13 & -32 & 4 \\ -6 & 29 & -12 \\ -20 & 8 & 49\end{array}\right)$
$ =\left(\begin{array}{ccc} 3+21-13 & -28+32 & -4 \\ -7+6 & 3+35-29 & -7+12 \\ -14+20 & -8 & 3+49-49 \end{array}\right)=\left(\begin{array}{ccc} 11 & 4 & -4 \\ -1 & 9 & 5 \\ 6 & -8 & 3 \end{array}\right) $
设 $\mathrm{P}=\left(\begin{array}{l}-3 \\ 2 \\ 5\end{array}\right), \mathrm{Q}=\left(\begin{array}{lll}4 & -2 & 3\end{array}\right), \mathrm{A}=\mathrm{PQ}$, 求 $\mathrm{A}^{100}$
解： 因 $Q P=(4,-2,3)\left(\begin{array}{c}-3 \\ 2 \\ 5\end{array}\right)=12-4+15=-1$
故 $A^{100}=(P Q)^{100}=P Q P Q P Q \cdots P Q$ $ =P[Q P]^{99} Q=(-1)^{99} P Q=-P Q $
由 $P Q=\left(\begin{array}{c}-3 \\ 2 \\ 5\end{array}\right)\left(\begin{array}{lll}4 & -2 & 3\end{array}\right)=\left(\begin{array}{ccc}-12 & 6 & -9 \\ 8 & -4 & 6 \\ 20 & -10 & 15\end{array}\right)$
得 $ A^{100}=-P Q=\left(\begin{array}{ccc} 12 & -6 & 9 \\ -8 & 4 & -6 \\ -20 & 10 & -15 \end{array}\right) $
求矩阵方程中的矩阵 $X:\left(\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right) X\left(\begin{array}{ccc}1 & -1 & 0 \\ -1 & 1 & 1 \\ 1 & 2 & 3\end{array}\right)=\left(\begin{array}{ccc}-6 & 3 & -2 \\ -10 & 7 & 3\end{array}\right)$
解: $X=\left(\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right)^{-1}\left(\begin{array}{ccc}-6 & 3 & -2 \\ -10 & 7 & 3\end{array}\right)\left(\begin{array}{ccc}1 & -1 & 0 \\ -1 & 1 & 1 \\ 1 & 2 & 3\end{array}\right)^{-1}$
$$ \begin{aligned} &=\left(\begin{array}{ll} -1 & 1 \\ -2 & 1 \end{array}\right)\left(\begin{array}{ccc} -6 & 3 & -2 \\ -10 & 7 & 3 \end{array}\right)\left(\begin{array}{ccc} \frac{-1}{3} & -1 & \frac{1}{3} \\ \frac{-4}{3} & -1 & \frac{1}{3} \\ 1 & 1 & 0 \end{array}\right) \\ &=\left(\begin{array}{ccc} -4 & 4 & 5 \\ 2 & 1 & 7 \end{array}\right)\left(\begin{array}{ccc} \frac{-1}{3} & -1 & \frac{1}{3} \\ \frac{-4}{3} & -1 & \frac{1}{3} \\ 1 & 1 & 0 \end{array}\right)=\left(\begin{array}{lll} 1 & 5 & 0 \\ 5 & 4 & 1 \end{array}\right) \end{aligned} $$求矩阵 $\left(\begin{array}{ccc}1 & 2 & -1 \\ 3 & 4 & -2 \\ 5 & -4 & 1\end{array}\right)$ 的逆阵
解: $\left(\begin{array}{ccc}1 & 2 & -1 \\ 3 & 4 & -2 \\ 5 & -4 & 1\end{array}\right)^{-1}=\frac{1}{2}\left(\begin{array}{ccc}-4 & 2 & 0 \\ -13 & 6 & -1 \\ -32 & 14 & -2\end{array}\right)=\left(\begin{array}{ccc}-2 & 1 & 0 \\ \frac{-13}{2} & 3 & \frac{-1}{2} \\ -16 & 7 & -1\end{array}\right)$
- 讨论对角矩阵 $\Lambda=\left(\begin{array}{llll}a_{1} & a_{2} & \cdots & a_{n}\end{array}\right)$ 的可逆性.
解: 当 $a_{1} a_{2} \cdots a_{n} \neq 0$ 时, $|\Lambda| \neq 0$. 此时, 矩阵 $\Lambda$ 可逆, 且: $\Lambda^{-1}=\left(\begin{array}{llll}\frac{1}{a_{1}} & & & \\ & \frac{1}{a_{2}} & & \\ & & \ddots & \\ & & & \frac{1}{a_{n}}\end{array}\right)$
当 $a_{1} a_{2} \cdots a_{n}=0$ 时, 矩阵 $\Lambda$ 不可逆.
设 $\mathbf{A}=\left(\begin{array}{ll}\mathbf{B} & \mathbf{D} \\ \mathbf{O} & \mathbf{C}\end{array}\right)$, 其中 $\mathrm{B} 、 \mathrm{C}$ 各为 $\mathrm{s}$ 阶和 $\mathrm{t}$ 阶可逆方阵, 证明 $\mathrm{A}$ 可逆, 并求 $\mathrm{A}^{-1}$.
证明:由 $|\mathbf{A}|=|\mathbf{B} \| \mathbf{C}|$. 因为 $\mathbf{B} 、 \mathbf{C}$ 可逆, 所以 $|\mathbf{A}| \neq 0$, 从 而 $A$ 是可逆方阵.
设 $\mathbf{A}^{-1}=\left(\begin{array}{ll}\mathbf{P} & \mathbf{R} \\ \mathbf{S} & \mathbf{Q}\end{array}\right)$, 得 $\mathbf{A} \mathbf{A}^{-1}=\left(\begin{array}{ll}\mathbf{B} & \mathbf{D} \\ \mathbf{O} & \mathbf{C}\end{array}\right)\left(\begin{array}{ll}\mathbf{P} & \mathbf{R} \\ \mathbf{S} & \mathbf{Q}\end{array}\right)$ $=\left(\begin{array}{cc}\mathbf{B P}+\mathbf{D S} & \mathbf{B R}+\mathbf{D R} \\ \mathbf{C S} & \mathbf{C} \mathbf{Q}\end{array}\right)=\left(\begin{array}{cc}\mathbf{E}_{s} & \mathbf{O} \\ \mathbf{O} & \mathbf{E}_{t}\end{array}\right)$
比较最后一个等式的两边得 $\mathbf{B R}+\mathbf{D Q}=\mathbf{O}$ $\mathbf{C Q}=\mathbf{E}_{t}$ CS $=0$ 在 $\mathrm{CQ}=\mathrm{E}$ 两边左乘 $\mathrm{C}^{-1}$ 得 $\mathrm{Q}=\mathrm{C}^{-1}$ 在 $\mathrm{CS}=0$ 两边左乘 $\mathrm{C}^{-1}$ 得 $\mathrm{S}=0$ $$ \mathbf{B P}+\mathbf{D S}=\mathbf{B P}=\mathbf{E}_{s} $$
在 $B P=E$ 两边左乘 $B^{-1}$ 得 $P=B^{-1}$ $B R+D Q=B R+D C^{-1}=O$ 或 $B R=-D C^{-1}$ 这个等式两边 左乘 $\mathbf{B}^{-1}$ 得 $\mathbf{R}=-\mathbf{B}^{-1} \mathbf{D C}^{-1}$ 结果为 $\mathbf{A}^{-1}=\left(\begin{array}{cc}\mathbf{B}^{-1} & -\mathbf{B}^{-1} \mathbf{D C}^{-1} \\ \mathbf{O} & \mathbf{C}^{-1}\end{array}\right)$