2次の行列と行列式随伴行列、複素数、積、等式

学習環境

行列と行列式 (現代数学への入門) (砂田利一(著)、岩波書店)の第1章(2次の行列と行列式)、演習問題1.14の解答を求めてみる。

A = [\begin{matrix} a & b \\ c & d \end{matrix}]

とおく。

\begin{array}{l} \det [\begin{matrix} a & b \\ c & d \end{matrix}] = 1 \\ a d - b c = 1 \end{array}

また、

\begin{array}{l} [\begin{matrix} \bar{a} & \bar{c} \\ \bar{b} & \bar{d} \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}] [\begin{matrix} a & b \\ c & d \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}] \\ [\begin{matrix} \bar{a} & \bar{c} \\ \bar{b} & \bar{d} \end{matrix}] [\begin{matrix} a & b \\ - c & - d \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}] \\ [\begin{matrix} {| a |}^{2} - {| c |}^{2} & \bar{a} b - \bar{c} d \\ a \bar{b} - c \bar{d} & {| b |}^{2} - {| d |}^{2} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}] \\ {| a |}^{2} - {| c |}^{2} = 1 \\ \bar{a} b - \bar{c} d = 0 \\ a \bar{b} - c \bar{d} = 0 \\ {| b |}^{2} - {| d |}^{2} = - 1 \end{array}

方程式を解く。

\begin{array}{l} a d \bar{b} - c {| d |}^{2} = 0 \\ (b c + 1) \bar{b} - c {| d |}^{2} = 0 \\ {| b |}^{2} c + \bar{b} - c {| d |}^{2} = 0 \\ c ({| b |}^{2} - {| d |}^{2}) = - \bar{b} \\ - c = - \bar{b} \\ c = \bar{b} \\ \bar{a} b c - {| c |}^{2} d = 0 \\ \bar{a} (a d - 1) - {| c |}^{2} d = 0 \\ {| a |}^{2} d - \bar{a} - {| c |}^{2} d = 0 \\ ({| a |}^{2} - {| c |}^{2}) d = \bar{a} \\ d = \bar{a} \end{array}

よって、

\begin{array}{l} {| a |}^{2} - {| b |}^{2} = {| a |}^{2} - {| \bar{c} |}^{2} = {| a |}^{2} - {| c |}^{2} = 1 \\ A = [\begin{matrix} a & b \\ \bar{b} & \bar{a} \end{matrix}] \end{array}