行列式行列式の幾何学的意味空間ベクトル、内積と外積、等式、行列式

学習環境

手を動かしてまなぶ線形代数 (藤岡敦(著)、裳華房)の第3章(行列式)、11(行列式の幾何学的意味)、基本問題の問11.4の解答を求めてみる。

\begin{array}{l} (a \times b) \cdot c \\ = \det [\begin{matrix} a \\ b \\ c \end{matrix}] \\ (b \times c) \cdot a \\ = \det [\begin{matrix} b \\ c \\ a \end{matrix}] \\ = - \det [\begin{matrix} c \\ b \\ a \end{matrix}] \\ = \det [\begin{matrix} a \\ b \\ c \end{matrix}] \\ (c \times a) \cdot b \\ = \det [\begin{matrix} c \\ a \\ b \end{matrix}] \\ = - \det [\begin{matrix} a \\ c \\ b \end{matrix}] \\ = \det [\begin{matrix} a \\ b \\ c \end{matrix}] \end{array}

よって、

\begin{array}{l} (a \times b) \cdot c \\ = (b \times c) \cdot a \\ = (c \times a) \cdot b \end{array}

\begin{array}{l} a = (a_{1}, a_{2}, a_{3}) \\ b = (b_{1}, b_{2}, b_{3}) \\ c = (c_{1}, c_{2}, c_{3}) \end{array}

とおく。

\begin{array}{l} a \times b = (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1}) \\ (a \times b) \times c \\ = ((a_{3} b_{1} - a_{1} b_{3}) c_{3} - (a_{1} b_{2} - a_{2} b_{1}) c_{2}, \\ (a_{1} b_{2} - a_{2} b_{1}) c_{1} - (a_{2} b_{3} - a_{3} b_{2}) c_{3}, \\ (a_{2} b_{3} - a_{3} b_{2}) c_{2} - (a_{3} b_{1} - a_{1} b_{3}) c_{1}) \\ = (a_{3} b_{1} c_{3} - a_{1} b_{3} c_{3} - a_{1} b_{2} c_{2} + a_{2} b_{1} c_{2}, \\ a_{1} b_{2} c_{1} - a_{2} b_{1} c_{1} - a_{2} b_{3} c_{3} + a_{3} b_{2} c_{3}, \\ a_{2} b_{3} c_{2} - a_{3} b_{2} c_{2} - a_{3} b_{1} c_{1} + a_{1} b_{3} c_{1}) \end{array}

また、

\begin{array}{l} (a \cdot c) b \\ = (a_{1} c_{1} + a_{2} c_{2} + a_{3} c_{3}) (b_{1}, b_{2}, b_{3}) \\ = (a_{1} b_{1} c_{1} + a_{2} b_{1} c_{2} + a_{3} b_{1} c_{3}, \\ a_{1} b_{2} c_{1} + a_{2} b_{2} c_{2} + a_{3} b_{2} c_{3}, \\ a_{1} b_{3} c_{1} + a_{2} b_{3} c_{2} + a_{3} b_{3} c_{3}) \\ (b \cdot c) a \\ = (b_{1} c_{1} + b_{2} c_{2} + b_{3} c_{3}) (a_{1}, a_{2}, a_{3}) \\ = (a_{1} b_{1} c_{1} + a_{1} b_{2} c_{2} + a_{1} b_{3} c_{3}, \\ a_{2} b_{1} c_{1} + a_{2} b_{2} c_{2} + a_{2} b_{3} c_{3}, \\ a_{3} b_{1} c_{1} + a_{3} b_{2} c_{2} + a_{3} b_{3} c_{3}) \\ (a \cdot c) b - (b \cdot c) a \\ = (a_{2} b_{1} c_{2} + a_{3} b_{1} c_{3} - a_{1} b_{2} c_{2} - a_{1} b_{3} c_{3}, \\ a_{1} b_{2} c_{1} + a_{3} b_{2} c_{3} - a_{2} b_{1} c_{1} - a_{2} b_{3} c_{3}, \\ a_{1} b_{3} c_{1} + a_{2} b_{3} c_{2} - a_{3} b_{1} c_{1} - a_{3} b_{2} c_{2}) \end{array}

よって、

(a \times b) \cdot c = (a \cdot c) b - (b \cdot c) a