「統計解析のための線形代数」復習筆記22

正方形矩陣 $A$ 的行列式滿足 $|A| \neq 0$ 時,逆矩陣可以表達爲(當 $|A|=0$ 時,正方形矩陣 $A$ 沒有逆矩陣): $$A^{-1}=\frac{1}{|A|}adj(A)=\frac{1}{|A|}(A_{ij})^t$$

$$=\frac{1}{|A|}\lbrace(-1)^{i+j}D_{ij}\rbrace^t$$

其中:

(1) 之前舉過的例子再拿來試試看:

$$X=\left( \begin{array}{c} 1 & 2 & 1 \newline 2 & 1 & 1 \newline 1 & 1 & 2 \end{array} \right)=\left(\begin{array}{c} x_{11} & x_{12} & x_{13} \newline x_{21} & x_{22} & x_{23} \newline x_{31} & x_{32} & x_{33} \end{array}\right)$$

元素 $x_{ij}$ 的餘因子 $X_{ij}(i,j=1,2,3)$ 爲:

$$X_{11}=(-1)^{1+1}\left| \begin{array}{c} 1 & 1 \newline 1 & 2 \end{array}\right|=1$$

$$X_{12}=(-1)^{1+2}\left| \begin{array}{c} 2 & 1 \newline 1 & 2 \end{array}\right|=-3$$

$$X_{13}=(-1)^{1+3}\left| \begin{array}{c} 2 & 1 \newline 1 & 1 \end{array}\right|=1$$

$$X_{21}=(-1)^{2+1}\left| \begin{array}{c} 2 & 1 \newline 1 & 2 \end{array}\right|=-3$$

$$X_{22}=(-1)^{2+2}\left| \begin{array}{c} 1 & 1 \newline 1 & 2 \end{array}\right|=1$$

$$X_{23}=(-1)^{2+3}\left| \begin{array}{c} 1 & 2 \newline 1 & 1 \end{array}\right|=1$$

$$X_{31}=(-1)^{3+1}\left| \begin{array}{c} 2 & 1 \newline 1 & 1 \end{array}\right|=1$$

$$X_{32}=(-1)^{3+2}\left| \begin{array}{c} 1 & 1 \newline 2 & 1 \end{array}\right|=1$$

$$X_{33}=(-1)^{3+3}\left| \begin{array}{c} 1 & 2 \newline 2 & 1 \end{array}\right|=-3$$

因此餘因子矩陣爲:$adj(X)=\left( \begin{array}{c} 1 & -3 & 1 \newline -3 & 1 & 1 \newline 1 & 1 & -3 \end{array} \right)^t=\left( \begin{array}{c} 1 & -3 & 1 \newline -3 & 1 & 1 \newline 1 & 1 & -3 \end{array} \right)$

我們看見這個餘因子矩陣是一個對稱矩陣,這是由於原矩陣 $X$ 本身就是一個對稱矩陣。另外,行列式爲:

$$\begin{align}|X|&=1\times X_{11}+2\times X_{12}+1\times X_{13}\newline&=1\times1+2\times(-3)+1\times1\newline&=-4\end{align}$$

因此所求的逆矩陣爲:

$$\begin{align}X^{-1}&=\frac{1}{|X|}adj(X)\newline &=\frac{1}{-4}\left( \begin{array}{c} 1 & -3 & 1 \newline -3 & 1 & 1 \newline 1 & 1 & -3 \end{array} \right)\newline &=\left( \begin{array}{c} -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} \newline \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} \newline -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} \end{array} \right)\end{align}$$

(2) 試求矩陣 $A=\left( \begin{array}{c} 1 & 2 & 1 \newline 2 & 3 & 1 \newline 1 & 2 & 2 \end{array} \right)=\left( \begin{array}{c} a_{11} & a_{12} & a_{13} \newline a_{21} & a_{22} & a_{23} \newline a_{31} & a_{32} & a_{33} \end{array} \right)$ 的逆矩陣 $A^{-1}$:

$$\begin{array} =A_{11}=6-2=4, & A_{12}=-(4-1)=-3, & A_{13}=4-3=1 \newline A_{21}=-(4-2)=-2, & A_{22}=2-1=1, & A_{23}=-(2-2)=0 \newline A_{31}=2-3=-1, & A_{32}=-(1-2)=1, & A_{33}=3-4=-1 \end{array}$$

$$adj(A)=\left( \begin{array}{c} 4 & -3 & 1 \newline -2 & 1 & 0 \newline -1 & 1 & -1 \end{array} \right)^t=\left( \begin{array}{c} 4 & -2 & -1 \newline -3 & 1 & 1 \newline 1 & 0 & -1 \end{array} \right)$$

$$\begin{align} |A| &=1\times A_{11}+2\times A_{12}+1\times A_{13} \newline &=1\times4+2\times(-3)+1\times1 \newline &=4-6+1 \newline &=-1 \end{align}$$

$$ \therefore \begin{align} A^{-1} &= \frac{1}{(-1)}\left( \begin{array}{c} 4 & -2 & -1 \newline -3 & 1 & 1 \newline 1 & 0 & -1 \end{array} \right) \newline &=\left( \begin{array}{c} -4 & 2 & 1 \newline 3 & -1 & -1 \newline -1 & 0 & 1 \end{array} \right) \end{align}$$

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