数学ガール　行列が描くもの　第3章の問題3-3　任意の2×2行列　A,B,C　に対して （AB)C＝A（BC） が成り立つことの証明。

$$A = \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ \end{pmatrix}$$

$$B = \begin{pmatrix} b_{11} & b_{12}\\ b_{21} & b_{22}\\ \end{pmatrix}$$

$$C = \begin{pmatrix} c_{11} & c_{12}\\ c_{21} & c_{22}\\ \end{pmatrix}$$

とする。

左辺(AB)C

$$\begin{pmatrix}\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ \end{pmatrix} \begin{pmatrix} b_{11} & b_{12}\\ b_{21} & b_{22}\\ \end{pmatrix}\end{pmatrix} \begin{pmatrix} c_{11} & c_{12}\\ c_{21} & c_{22}\\ \end{pmatrix}$$

$$= \begin{pmatrix} a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12}+ a_{12}b_{22} \\ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22} \\ \end{pmatrix} \begin{pmatrix} c_{11} & c_{12}\\ c_{21} & c_{22}\\ \end{pmatrix}$$

$$= \begin{pmatrix}\begin{pmatrix} a_{11}b_{11} + a_{12}b_{21}\end{pmatrix} c_{11} + \begin{pmatrix} a_{11}b_{12} + a_{12}b_{22}\end{pmatrix} c_{21} & \begin{pmatrix}a_{11}b_{11} + a_{12}b_{21}\end{pmatrix} c_{12} + \begin{pmatrix}a_{11}b_{12} + a_{12}b_{22}\end{pmatrix} c_{22} \\ \begin{pmatrix} a_{21}b_{11} + a_{22}b_{21}\end{pmatrix} c_{11} + \begin{pmatrix} a_{21}b_{12} + a_{22}b_{22}\end{pmatrix} c_{21} & \begin{pmatrix}a_{21}b_{11} + a_{22}b_{21}\end{pmatrix} c_{12} + \begin{pmatrix}a_{21}b_{12} + a_{22}b_{22}\end{pmatrix} c_{22} \end{pmatrix}$$

$$= \begin{pmatrix} a_{11}b_{11}c_{11} + a_{12}b_{21}c_{11} + a_{11}b_{12} c_{21} + a_{12}b_{22} c_{21} & a_{11}b_{11}c_{12} + a_{12}b_{21}c_{12} + a_{11}b_{12} c_{22} + a_{12}b_{22} c_{22} \\ a_{21}b_{11}c_{11} + a_{22}b_{21}c_{11} + a_{21}b_{12}c_{21} + a_{22}b_{22}c_{21} &a_{21}b_{11}c_{12} + a_{22}b_{21}c_{12} + a_{21}b_{12}c_{22} + a_{22}b_{22}c_{22} \end{pmatrix}$$

右辺A（BC)

$$\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ \end{pmatrix} \begin{pmatrix}\begin{pmatrix} b_{11} & b_{12}\\ b_{21} & b_{22}\\ \end{pmatrix} \begin{pmatrix} c_{11} & c_{12}\\ c_{21} & c_{22}\\ \end{pmatrix}\end{pmatrix}$$

$$= \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ \end{pmatrix} \begin{pmatrix} b_{11}c_{11} + b_{12}c_{21} & b_{11}c_{12}+ b_{12}c_{22} \\ b_{21}c_{11} + b_{22}c_{21} & b_{21}c_{12} + b_{22}c_{22} \\ \end{pmatrix}$$

$$= \begin{pmatrix} a_{11} \begin{pmatrix} b_{11}c_{11} + b_{12}c_{21}\end{pmatrix} + a_{12}\begin{pmatrix} b_{21}c_{11} + b_{22}c_{21}\end{pmatrix} & a_{11}\begin{pmatrix}b_{11}c_{12} + b_{12}c_{22}\end{pmatrix} + a_{12}\begin{pmatrix} b_{21}c_{12} + b_{22}c_{22}\end{pmatrix}\\ a_{21}\begin{pmatrix} b_{11}c_{11} + b_{12}c_{21}\end{pmatrix} + a_{22}\begin{pmatrix}b_{21}c_{11} + b_{22}c_{21}\end{pmatrix} & a_{21}\begin{pmatrix}b_{11}c_{12} + b_{12}c_{22}\end{pmatrix} + a_{22} \begin{pmatrix}b_{21}c_{12} + b_{22}c_{22}\end{pmatrix} \end{pmatrix}$$

$$= \begin{pmatrix} a_{11}b_{11}c_{11} + a_{11}b_{12}c_{21}+ a_{12}b_{21}c_{11} + a_{12}b_{22}c_{21} & a_{11}b_{11}c_{12} + a_{11}b_{12}c_{22} + a_{12}b_{21}c_{12} + a_{12}b_{22}c_{22}\\ a_{21}b_{11}c_{11} + a_{21}b_{12}c_{21} + a_{22}b_{21}c_{11} + a_{22}b_{22}c_{21} & a_{21}b_{11}c_{12} + a_{21}b_{12}c_{22}+ a_{22}b_{21}c_{12} + a_{22}b_{22}c_{22}\end{pmatrix}$$

左辺と右辺を比較

左辺

$$（AB)C= \begin{pmatrix} a_{11}b_{11}c_{11} + a_{12}b_{21}c_{11} + a_{11}b_{12} c_{21} + a_{12}b_{22} c_{21} & a_{11}b_{11}c_{12} + a_{12}b_{21}c_{12} + a_{11}b_{12} c_{22} + a_{12}b_{22} c_{22} \\ a_{21}b_{11}c_{11} + a_{22}b_{21}c_{11} + a_{21}b_{12}c_{21} + a_{22}b_{22}c_{21} &a_{21}b_{11}c_{12} + a_{22}b_{21}c_{12} + a_{21}b_{12}c_{22} + a_{22}b_{22}c_{22} \end{pmatrix}$$

右辺

$$A（BC）= \begin{pmatrix} a_{11}b_{11}c_{11} + a_{11}b_{12}c_{21} + a_{12}b_{21}c_{11} + a_{12}b_{22}c_{21} & a_{11}b_{11}c_{12} + a_{11}b_{12}c_{22} + a_{12}b_{21}c_{12} + a_{12}b_{22}c_{22} \\ a_{21}b_{11}c_{11} + a_{21}b_{12}c_{21} + a_{22}b_{21}c_{11} + a_{22}b_{22}c_{21} & a_{21}b_{11}c_{12} + a_{21}b_{12}c_{22}+ a_{22}b_{21}c_{12} + a_{22}b_{22}c_{22}\end{pmatrix}$$

計算の結果、成分が同一なので（AB)C=A(BC）が成り立つ。