余因子行列を用いた逆行列の計算方法

$$A=\left(\begin{array}{ccc}3& 1& 2\\ 5& 4& 1\\ 2& 1& 3\end{array}\right)$$

$$\begin{array}{rl}{A}_{11}& =11\\ \\ A_{12}& =-13\\ \\ A_{13}& =-3\\ \\ A_{21}& =-1\\ \\ A_{22}& =5\\ \\ A_{23}& =-1\\ \\ A_{31}& =-7\\ \\ A_{32}& =7\\ \\ A_{33}& =7\end{array}$$

$$\left(\begin{array}{ccc}11& -13& -3\\ -1& 5& -1\\ -7& 7& 7\end{array}\right)$$

$$\left(\begin{array}{ccc}11& -1& -7\\ -13& 5& 7\\ -3& -1& 7\end{array}\right)$$

$$\tilde{A}=\left(\begin{array}{ccc}11& -1& -7\\ -13& 5& 7\\ -3& -1& 7\end{array}\right)$$

$$\begin{array}{}\text{(1)}& A\tilde{A}=(detA)E\end{array}$$

$$\left|\begin{array}{ccc}3& 1& 2\\ 5& 4& 1\\ 2& 1& 3\end{array}\right|=14$$

$$\begin{array}{rl}A\tilde{A}& =\left(\begin{array}{ccc}3& 1& 2\\ 5& 4& 1\\ 2& 1& 3\end{array}\right)\left(\begin{array}{ccc}11& -1& -7\\ -13& 5& 7\\ -3& -1& 7\end{array}\right)\\ \\ & =\left(\begin{array}{ccc}14& 0& 0\\ 0& 14& 0\\ 0& 0& 14\end{array}\right)\end{array}$$

$$\frac{1}{detA}A\tilde{A}=E$$

$$A^{-1}=\frac{1}{detA}\tilde{A}$$

$$\begin{array}{rl}{A}^{-1}& =\frac{1}{detA}\tilde{A}\\ \\ & =\frac{1}{14}\left(\begin{array}{ccc}11& -1& -7\\ -13& 5& 7\\ -3& -1& 7\end{array}\right)\end{array}$$

$$X=\left(\begin{array}{ccc}-1& x+1& 1+{\displaystyle \frac{1}{x}}\\ 1+{\displaystyle \frac{1}{x}}& -1& x+1\\ x+1& 1+{\displaystyle \frac{1}{x}}& -1\end{array}\right)$$

$$X^{-1}=\frac{1}{detX}\tilde{X}$$

$$\begin{array}{rl}detX& =-1+(x+1{)}^3+{(1+\frac{1}{x})}^3-3(x+1)(1+\frac{1}{x})\\ \\ & =x^3+\frac{1}{x^3}+3x^2+\frac{3}{x^2}+6x+\frac{6}{x}+7\\ \\ & ={(x+\frac{1}{x})}^3+3x^2+\frac{3}{x^2}+3x+\frac{3}{x}+7\\ \\ & ={(x+\frac{1}{x})}^3+3{(x+\frac{1}{x})}^2+3x+\frac{3}{x}+1\\ \\ & ={(x+\frac{1}{x})}^3+3{(x+\frac{1}{x})}^2+3(x+\frac{1}{x})+1\\ \\ & ={(x+1+\frac{1}{x})}^3\end{array}$$

$$\begin{array}{rl}{X}_{11}& =\left|\begin{array}{cc}-1& x+1\\ 1+{\displaystyle \frac{1}{x}}& -1\end{array}\right|\\ \\ & =-x-1-\frac{1}{x}\\ \\ & =-(x+1+\frac{1}{x})\end{array}$$

$$\begin{array}{rl}{X}_{12}& =-\left|\begin{array}{cc}1+{\displaystyle \frac{1}{x}}& x+1\\ x+1& -1\end{array}\right|\\ \\ & =x^2+2x+2+{\displaystyle \frac{1}{x}}\\ \\ & =(x+1+\frac{1}{x})(x+1)\end{array}$$

$$\begin{array}{rl}{X}_{13}& =\left|\begin{array}{cc}1+{\displaystyle \frac{1}{x}}& -1\\ x+1& 1+{\displaystyle \frac{1}{x}}\end{array}\right|\\ \\ & =x+2+{\displaystyle \frac{2}{x}+{\displaystyle \frac{1}{x^2}}}\\ \\ & =(x+1+\frac{1}{x})(1+\frac{1}{x})\end{array}$$

$$\begin{array}{rl}{X}_{21}& =-\left|\begin{array}{cc}x+1& 1+{\displaystyle \frac{1}{x}}\\ 1+{\displaystyle \frac{1}{x}}& -1\end{array}\right|\\ \\ & =x+2+{\displaystyle \frac{2}{x}+{\displaystyle \frac{1}{x^2}}}\\ \\ & =(x+1+\frac{1}{x})(1+\frac{1}{x})\end{array}$$

$$\begin{array}{rl}{X}_{22}& =\left|\begin{array}{cc}-1& 1+{\displaystyle \frac{1}{x}}\\ x+1& -1\end{array}\right|\\ \\ & =-x-1-\frac{1}{x}\\ \\ & =-(x+1+\frac{1}{x})\end{array}$$

$$\begin{array}{rl}{X}_{23}& =-\left|\begin{array}{cc}-1& x+1\\ x+1& 1+{\displaystyle \frac{1}{x}}\end{array}\right|\\ \\ & =x^2+2x+2+{\displaystyle \frac{1}{x}}\\ \\ & =(x+1+\frac{1}{x})(x+1)\end{array}$$

$$\begin{array}{rl}{X}_{31}& =\left|\begin{array}{cc}x+1& 1+{\displaystyle \frac{1}{x}}\\ -1& x+1\end{array}\right|\\ \\ & =x^2+2x+2+{\displaystyle \frac{1}{x}}\\ \\ & =(x+1+\frac{1}{x})(x+1)\end{array}$$

$$\begin{array}{rl}{X}_{32}& =-\left|\begin{array}{cc}-1& 1+{\displaystyle \frac{1}{x}}\\ 1+{\displaystyle \frac{1}{x}}& x+1\end{array}\right|\\ \\ & =x+2+{\displaystyle \frac{2}{x}+{\displaystyle \frac{1}{x^2}}}\\ \\ & =(x+1+\frac{1}{x})(1+\frac{1}{x})\end{array}$$

$$\begin{array}{rl}{X}_{33}& =\left|\begin{array}{cc}-1& x+1\\ 1+{\displaystyle \frac{1}{x}}& -1\end{array}\right|\\ \\ & =-x-1-\frac{1}{x}\\ \\ & =-(x+1+\frac{1}{x})\end{array}$$

$$\begin{array}{rl}\tilde{X}& =\left(\begin{array}{ccc}{X}_{11}& X_{21}& X_{31}\\ X_{12}& X_{22}& X_{32}\\ X_{13}& X_{23}& X_{33}\end{array}\right)\\ \\ & =(x+1+{\displaystyle \frac{1}{x}})\left(\begin{array}{ccc}-1& 1+{\displaystyle \frac{1}{x}}& x+1\\ x+1& -1& 1+{\displaystyle \frac{1}{x}}\\ 1+{\displaystyle \frac{1}{x}}& x+1& -1\end{array}\right)\end{array}$$

$$\begin{array}{rl}{X}^{-1}& =\frac{1}{detX}\tilde{X}\\ \\ & ={\displaystyle \frac{1}{{(x+1+{\displaystyle \frac{1}{x}})}^2}\left(\begin{array}{ccc}-1& 1+{\displaystyle \frac{1}{x}}& x+1\\ x+1& -1& 1+{\displaystyle \frac{1}{x}}\\ 1+{\displaystyle \frac{1}{x}}& x+1& -1\end{array}\right)}\end{array}$$

$$A^{-1}=\frac{1}{detA}\tilde{A}$$