3次実正方行列の行列式
【問題】
$ 3次実正方行列
$ A = \left(\begin{array}{c}a_{11}~~a_{12}~~a_{13}\\a_{21}~~a_{22}~~a_{23}\\a_{31}~~a_{32}~~a_{33}\end{array}\right)
の行列式$ \det Aを，行列式の性質から導出せよ．
【解答】
$ \det A=\det(\bold{a_1, a_2, a_3})
$ ~~~~~~~~~~=\det(a_{11}\bold{e_1}+a_{21}\bold{e_2}+a_{31}\bold{e_3}, \bold{a_2}, \bold{a_3})
$ ~~~~~~~~~~=a_{11}\det(\bold{e_1}, \bold{a_2, a_3})+a_{21}\det(\bold{e_2}, \bold{a_2, a_3})+a_{31}\det(\bold{e_3}, \bold{a_2, a_3})
$ ~~~~~~~~~\because列ベクトル$ \bold{a_1}を線形結合で表現し，行列式の多重線形性により展開．
$ ~~~~~~~~~~=a_{11}\det(\bold{e_1}, a_{12}\bold{e_1}+a_{22}\bold{e_2}+a_{32}\bold{e_3}, \bold{a_3})
$ ~~~~~~~~~+a_{21}\det(\bold{e_2}, a_{12}\bold{e_1}+a_{22}\bold{e_2}+a_{32}\bold{e_3}, \bold{a_3})
$ ~~~~~~~~~+a_{31}\det(\bold{e_3}, a_{12}\bold{e_1}+a_{22}\bold{e_2}+a_{32}\bold{e_3}, \bold{a_3})
$ ~~~~~~~~~\because列ベクトル$ \bold{a_2}を線形結合で表現し，行列式の多重線形性により展開．
$ ~~~~~~~~~~=a_{11}a_{22}\det(\bold{e_1}, \bold{e_2},\bold{a_3})+a_{11}a_{32}\det(\bold{e_1}, \bold{e_3},\bold{a_3})
$ ~~~~~~~~~~+a_{21}a_{12}\det(\bold{e_2}, \bold{e_1},\bold{a_3})+a_{21}a_{32}\det(\bold{e_2}, \bold{e_3},\bold{a_3})
$ ~~~~~~~~~~+a_{31}a_{12}\det(\bold{e_3}, \bold{e_1},\bold{a_3})+a_{31}a_{22}\det(\bold{e_3}, \bold{e_2},\bold{a_3})
$ ~~~~~~~~~\because同じ単位ベクトル$ \bold{e_i}が含まれる項は0になる．
$ ~~~~~~~~~~=a_{11}a_{22}\det(\bold{e_1}, \bold{e_2},a_{13}\bold{e_1}+a_{23}\bold{e_2}+a_{33}\bold{e_3})+a_{11}a_{32}\det(\bold{e_1}, \bold{e_3},a_{13}\bold{e_1}+a_{23}\bold{e_2}+a_{33}\bold{e_3})
$ ~~~~~~~~~~+a_{21}a_{12}\det(\bold{e_2}, \bold{e_1},a_{13}\bold{e_1}+a_{23}\bold{e_2}+a_{33}\bold{e_3})+a_{21}a_{32}\det(\bold{e_2}, \bold{e_3},a_{13}\bold{e_1}+a_{23}\bold{e_2}+a_{33}\bold{e_3})
$ ~~~~~~~~~~+a_{31}a_{12}\det(\bold{e_3}, \bold{e_1},a_{13}\bold{e_1}+a_{23}\bold{e_2}+a_{33}\bold{e_3})+a_{31}a_{22}\det(\bold{e_3}, \bold{e_2},a_{13}\bold{e_1}+a_{23}\bold{e_2}+a_{33}\bold{e_3})
$ ~~~~~~~~~\because列ベクトル$ \bold{a_3}を線形結合で表現し，行列式の多重線形性により展開．
$ ~~~~~~~~~~=a_{11}a_{22}a_{33}\det(\bold{e_1}, \bold{e_2},\bold{e_3})+a_{11}a_{32}a_{23}\det(\bold{e_1}, \bold{e_3},\bold{e_2})+a_{21}a_{12}a_{33}\det(\bold{e_2}, \bold{e_1},\bold{e_3})
$ ~~~~~~~~~~+a_{21}a_{32}a_{13}\det(\bold{e_2}, \bold{e_3},\bold{e_1})+a_{31}a_{12}a_{23}\det(\bold{e_3}, \bold{e_1},\bold{e_2})+a_{31}a_{22}a_{13}\det(\bold{e_3}, \bold{e_2},\bold{e_1})
$ ~~~~~~~~~\because同じ単位ベクトル$ \bold{e_i}が含まれる項は0になる．
$ ~~~~~~~~~~=a_{11}a_{22}a_{33}\det(\bold{e_1}, \bold{e_2},\bold{e_3})-a_{11}a_{32}a_{23}\det(\bold{e_1}, \bold{e_2},\bold{e_3})-a_{21}a_{12}a_{33}\det(\bold{e_1}, \bold{e_2},\bold{e_3})
$ ~~~~~~~~~~+a_{21}a_{32}a_{13}\det(\bold{e_1}, \bold{e_2},\bold{e_3})+a_{31}a_{12}a_{23}\det(\bold{e_1}, \bold{e_2},\bold{e_3})-a_{31}a_{22}a_{13}\det(\bold{e_1}, \bold{e_2},\bold{e_3})
$ ~~~~~~~~~\becauseベクトルの交代性により，$ \bold{e_i},\bold{e_j}を1回入れ替えると，符号反転．
$ ~~~~~~~~~~=a_{11}a_{22}a_{33}-a_{11}a_{32}a_{23}-a_{21}a_{12}a_{33}+a_{21}a_{32}a_{13}+a_{31}a_{12}a_{23}-a_{31}a_{22}a_{13}//
$ ~~~~~~~~~\because$ \det(\bold{e_1},\bold{e_2},\bold{e_3})=\det{E_3}=1
【補足】
3次正方行列の行列式を求める公式として，サラスの公式がある．