【線形代数】4-5-1 余因子展開｜問題集
1.次の行列式の値を求めなさい。
(1)\(A=\begin{vmatrix}a & -b & -c & -d \\ b & a & -d & c \\ c & d & a & -b \\ d & -c & b & a\end{vmatrix}\)
\(=(-1)^2a\begin{vmatrix}a & -d & c \\ d & a & -b \\ -c & b & a\end{vmatrix}+(-1)^3b\begin{vmatrix}-b & -c & -d \\ d & a & -b \\ -c & b & a\end{vmatrix}\)
\(\ \ \ +(-1)^4c\begin{vmatrix}-b & -c & -d \\ a & -d & c \\ -c & b & a\end{vmatrix}+(-1)^5d\begin{vmatrix}-b & -c & -d \\ a & -d & c \\ d & a & -b\end{vmatrix}\)
\(=a^2(a^2+b^2+c^2+d^2)+b^2(a^2+b^2+c^2+d^2)\)
\(\ \ \ +c^2(a^2+b^2+c^2+d^2)+d^2(a^2+b^2+c^2+d^2)\)
\(=(a^2+b^2+c^2+d^2)^2\)
\(\ \ \ +(-1)^4c\begin{vmatrix}-b & -c & -d \\ a & -d & c \\ -c & b & a\end{vmatrix}+(-1)^5d\begin{vmatrix}-b & -c & -d \\ a & -d & c \\ d & a & -b\end{vmatrix}\)
\(=a^2(a^2+b^2+c^2+d^2)+b^2(a^2+b^2+c^2+d^2)\)
\(\ \ \ +c^2(a^2+b^2+c^2+d^2)+d^2(a^2+b^2+c^2+d^2)\)
\(=(a^2+b^2+c^2+d^2)^2\)
(2)\(A=\begin{vmatrix}\sin\theta\cos\varphi & r\cos\theta\cos\varphi & -r\sin\theta\sin\varphi \\ \sin\theta\sin\varphi & r\cos\theta\sin\varphi & r\sin\theta\cos\varphi \\ \cos\theta & -r\sin\theta & 0\end{vmatrix}\)
\(=(-1)^4\cos\theta\begin{vmatrix}r\cos\theta\cos\varphi & -r\sin\theta\sin\varphi \\ r\cos\theta\sin\varphi & r\sin\theta\cos\varphi \end{vmatrix}\)
\(\ \ \ +(-1)^5(-r\sin\theta)\begin{vmatrix}\sin\theta\cos\varphi & -r\sin\theta\sin\varphi \\ \sin\theta\sin\varphi & r\sin\theta\cos\varphi \end{vmatrix}\)
\(=r^2\sin\theta\cos^2\theta\begin{vmatrix}\cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \end{vmatrix}\)
\(\ \ \ +r^2\sin^3\theta\begin{vmatrix}\cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \end{vmatrix}\)
\(=(r^2\sin\theta\cos^2\theta+r^2\sin^3\theta)(\cos^2\varphi+\sin^2\varphi)\)
\(=r^2\sin\theta(\cos^2\theta+\sin^2\theta)\)
\(=r^2\sin\theta\)
\(\ \ \ +(-1)^5(-r\sin\theta)\begin{vmatrix}\sin\theta\cos\varphi & -r\sin\theta\sin\varphi \\ \sin\theta\sin\varphi & r\sin\theta\cos\varphi \end{vmatrix}\)
\(=r^2\sin\theta\cos^2\theta\begin{vmatrix}\cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \end{vmatrix}\)
\(\ \ \ +r^2\sin^3\theta\begin{vmatrix}\cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \end{vmatrix}\)
\(=(r^2\sin\theta\cos^2\theta+r^2\sin^3\theta)(\cos^2\varphi+\sin^2\varphi)\)
\(=r^2\sin\theta(\cos^2\theta+\sin^2\theta)\)
\(=r^2\sin\theta\)
(3)\(A=\begin{vmatrix}x^2+1 & x & 0 & \cdots & 0 & 0 \\ x & x^2+1 & x & \cdots & 0 & 0 \\ 0 & x & x^2+1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & x^2+1 & x \\ 0 & 0 & 0 & \cdots & x & x^2+1\end{vmatrix}\)
求めるn次行列式を\(f_n(x)\)とおく。
\(f_n(x)=(x^2+1)\begin{vmatrix}x^2+1 & x & 0 & 0 & 0 \\ x & x^2+1 & x & 0 & 0 \\ 0 & \ddots & \ddots & \ddots & 0 \\ 0 & 0 & x & x^2+1 & x \\ 0 & 0 & 0 & x & x^2+1\end{vmatrix}\)
\(\ \ \ \ \ -x\begin{vmatrix}x & 0 & 0 & 0 & 0 \\ x & x^2+1 & x & 0 & 0 \\ 0 & \ddots & \ddots & \ddots & 0 \\ 0 & 0 & x & x^2+1 & x \\ 0 & 0 & 0 & x & x^2+1\end{vmatrix}\)
\(=(x^2+1)f_{n-1}(x)-x^2\begin{vmatrix}x^2+1 & x & 0 & 0 \\ 0 & \ddots & \ddots & 0 \\ 0 & x & x^2+1 & x \\ 0 & 0 & x & x^2+1\end{vmatrix}\)
\(=(x^2+1)f_{n-1}(x)-x^2f_{n-2}(x)\)
すなわち
\(f_{n+2}(x)=(x^2+1)f_{n+1}(x)-x^2f_n(x)\)
\(f_{n+2}(x)-f_{n+1}(x)=x^2\{f_{n+1}(x)-f_n(x)\}\)
また、
\(f_1(x)=x^2+1\)
\(f_2(x)=\begin{vmatrix}x^2+1 & x \\ x & x^2+1 \end{vmatrix}=x^4+x^2+1\)
なので
\(f_{n+1}-f_n(x)=(x^2)^{n-1}\{f_2(x)-f_1(x)\}=x^{2n+2}\)
よって、
\(\displaystyle f_n(x)=f_1(x)+\sum_{k=1}^{n-1}x^{2k+2}\)
\(\displaystyle =(x^2+1)+\sum_{k=2}^{n}x^{2k}\)
\(\displaystyle =\sum_{k=0}^{n}x^{2k}\)
\(f_n(x)=(x^2+1)\begin{vmatrix}x^2+1 & x & 0 & 0 & 0 \\ x & x^2+1 & x & 0 & 0 \\ 0 & \ddots & \ddots & \ddots & 0 \\ 0 & 0 & x & x^2+1 & x \\ 0 & 0 & 0 & x & x^2+1\end{vmatrix}\)
\(\ \ \ \ \ -x\begin{vmatrix}x & 0 & 0 & 0 & 0 \\ x & x^2+1 & x & 0 & 0 \\ 0 & \ddots & \ddots & \ddots & 0 \\ 0 & 0 & x & x^2+1 & x \\ 0 & 0 & 0 & x & x^2+1\end{vmatrix}\)
\(=(x^2+1)f_{n-1}(x)-x^2\begin{vmatrix}x^2+1 & x & 0 & 0 \\ 0 & \ddots & \ddots & 0 \\ 0 & x & x^2+1 & x \\ 0 & 0 & x & x^2+1\end{vmatrix}\)
\(=(x^2+1)f_{n-1}(x)-x^2f_{n-2}(x)\)
すなわち
\(f_{n+2}(x)=(x^2+1)f_{n+1}(x)-x^2f_n(x)\)
\(f_{n+2}(x)-f_{n+1}(x)=x^2\{f_{n+1}(x)-f_n(x)\}\)
また、
\(f_1(x)=x^2+1\)
\(f_2(x)=\begin{vmatrix}x^2+1 & x \\ x & x^2+1 \end{vmatrix}=x^4+x^2+1\)
なので
\(f_{n+1}-f_n(x)=(x^2)^{n-1}\{f_2(x)-f_1(x)\}=x^{2n+2}\)
よって、
\(\displaystyle f_n(x)=f_1(x)+\sum_{k=1}^{n-1}x^{2k+2}\)
\(\displaystyle =(x^2+1)+\sum_{k=2}^{n}x^{2k}\)
\(\displaystyle =\sum_{k=0}^{n}x^{2k}\)
次の学習に進もう！