求解逆矩阵的三种方法

一、题目

设矩阵 $\boldsymbol{A} = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$, $\boldsymbol{B} = \boldsymbol{A}^{2} – 3 \boldsymbol{A} + 2 \boldsymbol{E}$, 则 $\boldsymbol{B}^{-1} = \underline{\quad \quad \quad}$

二、解析

首先：

$$
\begin{aligned}
\textcolor{lightgreen}{ \boldsymbol{B} } & = \boldsymbol{A}^{2} – 3 \boldsymbol{A} + 2 \boldsymbol{E} \\ \\
& = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}^{2} – 3 \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ \\
& = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} – 3 \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ \\
& = \begin{bmatrix} -1 & -4 \\ 8 & 7 \end{bmatrix} + \begin{bmatrix} -3 & 3 \\ -6 & -9 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \\ \\
& = \textcolor{lightgreen}{ \begin{bmatrix} -2 & -1 \\ 2 & 0 \end{bmatrix} }
\end{aligned}
$$

求解 $\boldsymbol{B}^{-1}$ 的方法 1：初等行变换

根据《初等变换求逆法的形象理解：把单位矩阵 E 看作一张“白纸”或“原点”》这篇文章可知：

$$
\begin{bmatrix} \boldsymbol{A}, \boldsymbol{E} \end{bmatrix} \textcolor{lightgreen}{ \leadsto } \text{初等行变换} \textcolor{lightgreen}{ \leadsto } \begin{bmatrix} \boldsymbol{E}, \boldsymbol{A}^{-1} \end{bmatrix}
$$

因此，由 $\boldsymbol{B}$ $=$ $\begin{bmatrix} -2 & -1 \\ 2 & 0 \end{bmatrix}$ 可得：

$$
\begin{aligned}
& \left[\begin{array}
{cc|cc}
-2 & -1 & 1 & 0 \\
2 & 0 & 0 & 1
\end{array}\right] \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \left[\begin{array}
{cc|cc}
-2 & -1 & 1 & 0 \\
1 & 0 & 0 & \frac{1}{2}
\end{array}\right] \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \left[\begin{array}
{cc|cc}
1 & 0 & 0 & \frac{1}{2} \\
-2 & -1 & 1 & 0
\end{array}\right] \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \left[\begin{array}
{cc|cc}
1 & 0 & 0 & \frac{1}{2} \\
0 & -1 & 1 & 1
\end{array}\right] \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \left[\begin{array}
{cc|cc}
1 & 0 & \textcolor{lightgreen}{0} & \textcolor{lightgreen}{\frac{1}{2}} \\
0 & 1 & \textcolor{lightgreen}{-1} & \textcolor{lightgreen}{-1}
\end{array}\right]
\end{aligned}
$$

于是可知：

$$
\textcolor{lightgreen}{ \boldsymbol{B}^{-1} = \begin{bmatrix} 0 & \frac{1}{2} \\ -1 & -1 \end{bmatrix} }
$$

求解 $\boldsymbol{B}^{-1}$ 的方法 2：初等列变换

根据《初等变换求逆法的形象理解：把单位矩阵 E 看作一张“白纸”或“原点”》这篇文章可知：

$$
\begin{bmatrix} \boldsymbol{A} \\ \boldsymbol{E} \end{bmatrix} \textcolor{lightgreen}{ \leadsto } \text{初等列变换} \textcolor{lightgreen}{ \leadsto } \begin{bmatrix} \boldsymbol{E} \\ \boldsymbol{A}^{-1} \end{bmatrix}
$$

因此，由 $\boldsymbol{B}$ $=$ $\begin{bmatrix} -2 & -1 \\ 2 & 0 \end{bmatrix}$ 可得：

$$
\begin{aligned}
& \left[\begin{array}
{cc}
-2 & -1 \\
2 & 0 \\
\hline
1 & 0 \\
0 & 1
\end{array}\right] \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \left[\begin{array}
{cc}
-1 & -2 \\
0 & 2 \\
\hline
0 & 1 \\
1 & 0
\end{array}\right] \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \left[\begin{array}
{cc}
1 & -2 \\
0 & 2 \\
\hline
0 & 1 \\
-1 & 0
\end{array}\right] \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \left[\begin{array}
{cc}
1 & -1 \\
0 & 1 \\
\hline
0 & \frac{1}{2} \\
-1 & 0
\end{array}\right] \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \left[\begin{array}
{cc}
1 & 0 \\
0 & 1 \\
\hline
\textcolor{lightgreen}{0} & \textcolor{lightgreen}{\frac{1}{2}} \\
\textcolor{lightgreen}{-1} & \textcolor{lightgreen}{-1}
\end{array}\right]
\end{aligned}
$$

于是可知：

$$
\textcolor{lightgreen}{ \boldsymbol{B}^{-1} = \begin{bmatrix} 0 & \frac{1}{2} \\ -1 & -1 \end{bmatrix} }
$$

求解 $\boldsymbol{B}^{-1}$ 的方法 3：伴随矩阵

由于本题中的矩阵 $\boldsymbol{B}$ 是一个二阶矩阵，且：

$$
\boldsymbol{B}^{*} = |\boldsymbol{B}| \boldsymbol{B}^{-1} \textcolor{lightgreen}{ \leadsto } \boldsymbol{B}^{-1} = \frac{\boldsymbol{B}^{*}}{|\boldsymbol{B}|}
$$

因此，由《快速求解二阶矩阵的伴随矩阵》和 $\boldsymbol{B}$ $=$ $\begin{bmatrix} -2 & -1 \\ 2 & 0 \end{bmatrix}$ 可知：

$$
\boldsymbol{B}^{*} = \begin{bmatrix} 0 & 1 \\ -2 & -2 \end{bmatrix}
$$

于是：

$$
\textcolor{lightgreen}{ \boldsymbol{B}^{-1} } = \frac{\boldsymbol{B}^{*}}{|\boldsymbol{B}|} = \dfrac{1}{2} \begin{bmatrix} 0 & 1 \\ -2 & -2 \end{bmatrix} = \textcolor{lightgreen}{ \begin{bmatrix} 0 & \frac{1}{2} \\ -1 & -1 \end{bmatrix} }
$$

由于二阶矩阵可以使用公式快速计算出伴随矩阵，而求解三阶及以上阶数矩阵的伴随矩阵就变得很麻烦，所以，这里的方法 3 主要适用于二阶矩阵的逆矩阵求解.

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