【详解】方法 1: 由 $A A^{*}=A^{*} A=|A| E$, 知 $\left|A^{*}\right|=|A|^{n-1}$, 因此有 $8=\left|A^{*}\right|=|A|^{3}$,

\begin{aligned} \text { 化简 } & \Rightarrow|A| B E=A^{*} B E+3 A^{*} A \Rightarrow 2 B=A^{*} B+3|A| E \\ & \Rightarrow 2 B=A^{*} B+6 E \Rightarrow\left(2 E-A^{*}\right) B=6 E \end{aligned}

\begin{aligned} B &=\left(2 E-A^{*}\right)^{-1} \\ &=6\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & -6 \end{array}\right]^{-1}=6\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & \frac{1}{2} & 0 & -\frac{1}{6} \end{array}\right]=\left[\begin{array}{cccc} 6 & 0 & 0 & 0 \\ 0 & 6 & 0 & 0 \\ 6 & 0 & 6 & 0 \\ 0 & 3 & 0 & -1 \end{array}\right] \end{aligned}
(由初等变换法求得)

$$A \mid\left(A^{*}\right)^{-1}=2\left(A^{*}\right)^{-1}=2\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & \frac{3}{8} & 0 & \frac{1}{8} \end{array}\right]=\left[\begin{array}{cccc} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ -2 & 0 & 2 & 0 \\ 0 & \frac{3}{4} & 0 & \frac{1}{4} \end{array}\right] \text {, }$$

(由初等变换法求得), 可见 $A-E$ 为逆矩阵.

$$(A-E)^{-1}=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -2 & 0 & 1 & 0 \\ 0 & \frac{3}{4} & 0 & -\frac{3}{4} \end{array}\right]^{-1}=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 2 & 0 & 1 & 0 \\ 0 & 1 & 0 & -\frac{3}{4} \end{array}\right]$$

$$B=3\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 2 & 0 & 1 & 0 \\ 0 & 1 & 0 & -\frac{3}{4} \end{array}\right]\left[\begin{array}{cccc} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ -2 & 0 & 2 & 0 \\ 0 & \frac{3}{4} & 0 & \frac{1}{4} \end{array}\right]=\left[\begin{array}{cccc} 6 & 0 & 0 & 0 \\ 0 & 6 & 0 & 0 \\ 6 & 0 & 6 & 0 \\ 0 & 3 & 0 & -1 \end{array}\right]$$

$$B=3(A-E)^{-1} A=3\left[A^{-1}(A-E)\right]^{-1}=3\left(E-A^{-1}\right)^{-1}=3\left(E-\frac{A^{*}}{|A|}\right)^{-1}$$

$$B=3\left(E-\frac{A^{*}}{2}\right)^{-1}=3 \cdot\left(\frac{2 E-A^{*}}{2}\right)^{-1}=6\left(2 E-A^{*}\right)^{-1}$$

$$2 E-A^{*}=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & -6 \end{array}\right],\left(2 E-A^{*}\right)^{-1}=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & \frac{1}{2} & 0 & -\frac{1}{6} \end{array}\right] \text {, }$$

$$B=6\left(2 E-A^{*}\right)^{-1}=6\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & -6 \end{array}\right]=\left[\begin{array}{cccc} 6 & 0 & 0 & 0 \\ 0 & 6 & 0 & 0 \\ 6 & 0 & 6 & 0 \\ 0 & 3 & 0 & -1 \end{array}\right]$$
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