数学 - オイラーの公式とその応用(Euler's Formula & Its Applications - ベクトルと行列(Vector & Matrix) - 逆行列と連立一次方程式の解法(行列式) - オイラーの贈物

オイラーの贈物

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オイラーの贈物―人類の至宝eiπ=-1を学ぶ (吉田 武(著)、東海大学出版会)の第2部(オイラーの公式とその応用(Euler's Formula & Its Applications)、第9章(ベクトルと行列(Vector & Matrix))、9.3(逆行列と連立一次方程式の解法)、問題5.を取り組んでみる。

問題5.

$\begin{array}{l}\det D=\sqrt{2}\sqrt{3}-1=\sqrt{6}-1\\ \\ D^{-1}=\frac{1}{\sqrt{6}-1}\left(\begin{array}{cc}\sqrt{3}& -1\\ -1& \sqrt{2}\end{array}\right)\\ \frac{1}{\sqrt{6}-1}\left(\begin{array}{cc}\sqrt{3}& -1\\ -1& \sqrt{2}\end{array}\right)\left(\begin{array}{cc}\sqrt{2}& 1\\ 1& \sqrt{3}\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)\\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)\\ =\left(\begin{array}{l}x\\ y\end{array}\right)\\ \\ \frac{1}{\sqrt{6}-1}\left(\begin{array}{cc}\sqrt{3}& -1\\ -1& \sqrt{2}\end{array}\right)\left(\begin{array}{l}2+\sqrt{3}\\ 3+\sqrt{2}\end{array}\right)\\ =\frac{1}{\sqrt{6}-1}\left(\begin{array}{l}2\sqrt{3}+3-3-\sqrt{2}\\ -2-\sqrt{3}+3\sqrt{2}+2\end{array}\right)\\ =\frac{1}{\sqrt{6}-1}\left(\begin{array}{l}2\sqrt{3}-\sqrt{2}\\ 3\sqrt{2}-\sqrt{3}\end{array}\right)\\ =\left(\begin{array}{l}\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(\sqrt{6}+1\right)}{5}\\ \frac{\left(3\sqrt{2}-\sqrt{3}\right)\left(\sqrt{6}+1\right)}{5}\end{array}\right)\\ =\left(\begin{array}{l}\frac{6\sqrt{2}+2\sqrt{3}-2\sqrt{3}-\sqrt{2}}{5}\\ \frac{6\sqrt{3}+3\sqrt{2}-3\sqrt{2}-\sqrt{3}}{5}\end{array}\right)\\ =\left(\begin{array}{l}\sqrt{2}\\ \sqrt{3}\end{array}\right)\\ x=\sqrt{2},y=\sqrt{3}\end{array}$