Example 5.23

Given

$$\mathbf{A} = \begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix} \quad \text{and} \quad \mathbf{B} = \begin{bmatrix} 0 & 1 \ 1 & 1 \end{bmatrix}$$

evaluate $(\mathbf{AB})^{-1}$ , $\mathbf{A}^{-1}\mathbf{B}^{-1}$ , $\mathbf{B}^{-1}\mathbf{A}^{-1}$ 并证明 $(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$ .

Solution

$$\mathbf{A}^{-1} = \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} \ \frac{2}{3} & -\frac{1}{3} \end{bmatrix}, \quad \mathbf{B}^{-1} = \begin{bmatrix} -1 & 1 \ 1 & 0 \end{bmatrix}$$

$$\mathbf{AB} = \begin{bmatrix} 2 & 3 \ 1 & 3 \end{bmatrix}, \quad (\mathbf{AB})^{-1} = \begin{bmatrix} 1 & -1 \ -\frac{1}{3} & \frac{2}{3} \end{bmatrix}$$

$$\mathbf{A}^{-1}\mathbf{B}^{-1} = \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} \ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} -1 & 1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{1}{3} \ -1 & \frac{2}{3} \end{bmatrix}$$

$$\mathbf{B}^{-1}\mathbf{A}^{-1} = \begin{bmatrix} -1 & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix} -\frac{1}{3} & \frac{2}{3} \ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} = \begin{bmatrix} 1 & -1 \ -\frac{1}{3} & \frac{2}{3} \end{bmatrix} = (\mathbf{AB})^{-1}$$

Example 5.24

给定两个矩阵

$$\mathbf{A} = \begin{bmatrix} 0 & -\frac{3}{5} & 0 \ \frac{5}{3} & 0 & -\frac{5}{3} \ 0 & 6 & -6 \end{bmatrix} \quad \text{and} \quad \mathbf{T} = \begin{bmatrix} 0.6 & 0.3 & 0.1 \ 1 & 1 & 0.5 \ 1.2 & 1.5 & 1 \end{bmatrix}$$

证明矩阵 $\mathbf{T}^{-1}\mathbf{AT}$ 是对角的.

Solution

逆矩阵最好用 MATLAB 或类似软件包来计算。. 通过直接相乘可以验证

$$\mathbf{T}^{-1} = \frac{1}{6} \begin{bmatrix} 25 & -15 & 5 \ -40 & 48 & -20 \ 30 & -54 & 30 \end{bmatrix}$$

继续相乘可得

$$\frac{1}{6} \begin{bmatrix} 25 & -15 & 5 \ -40 & 48 & -20 \ 30 & -54 & 30 \end{bmatrix} \begin{bmatrix} 0 & -\frac{3}{5} & 0 \ \frac{5}{3} & 0 & -\frac{5}{3} \ 0 & 6 & -6 \end{bmatrix} \begin{bmatrix} 0.6 & 0.3 & 0.1 \ 1 & 1 & 0.5 \ 1.2 & 1.5 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \ 0 & -2 & 0 \ 0 & 0 & -3 \end{bmatrix}$$