行列式の積(乗法定理)の証明(2次)

2次正方行列  A ,  B に対して,
 \begin{vmatrix}AB\end{vmatrix}=\begin{vmatrix}A\end{vmatrix}\begin{vmatrix}B\end{vmatrix}
を証明せよ。

(解答)
 A=\begin{pmatrix}
                                          a_{11}&a_{12}\\a_{21}&a_{22}
                                       \end{pmatrix}
                                      =
                                        \begin{pmatrix}\overrightarrow{a}_{1}&\overrightarrow{a}_{2}
                                        \end{pmatrix} ,  B=\begin{pmatrix}
                                      b_{11}&b_{12}\\b_{21}&b_{22}
                                    \end{pmatrix}
とすると、
 AB = \left(
    \begin{array}{ccc}
      b_{11}\overrightarrow{a}_{1}+b_{21}\overrightarrow{a}_{2}&b_{12}\overrightarrow{a}_{1}+b_{22}\overrightarrow{a}_{2}
    \end{array}
  \right)
よって,
|AB| = \left|
    \begin{array}{cc}
      b_{11}\overrightarrow{a}_{1}+b_{21}\overrightarrow{a}_{2}&b_{12}\overrightarrow{a}_{1}+b_{22}\overrightarrow{a}_{2}
    \end{array}
   \right|
 =  b_{11} \left|
    \begin{array}{cc}
      \overrightarrow{a}_{1}&b_{12}\overrightarrow{a}_{1}+b_{22}\overrightarrow{a}_{2}
    \end{array}
   \right|  +b_{21} \left|
    \begin{array}{ccc}
      \overrightarrow{a}_{2}&b_{12}\overrightarrow{a}_{1}+b_{22}\overrightarrow{a}_{2}
    \end{array}
   \right| (注1)
 = b_{11}b_{12}\begin{vmatrix}\overrightarrow{a}_1&\overrightarrow{a}_1\end{vmatrix}+ b_{11}b_{22}\begin{vmatrix}\overrightarrow{a}_1&\overrightarrow{a}_2\end{vmatrix}  + b_{21}b_{12}\begin{vmatrix}\overrightarrow{a}_2&\overrightarrow{a}_1\end{vmatrix}+ b_{21}b_{22}\begin{vmatrix}\overrightarrow{a}_2&\overrightarrow{a}_2\end{vmatrix}   (注2)
 = \begin{vmatrix}
                \overrightarrow{a}_1&\overrightarrow{a}_2
              \end{vmatrix}(
                                b_{11}b_{22}-b_{21}b_{12}
                            )    (注3)
 = |A||B|

(注)
1.第1列について、行列式の線形性から。
2.各項の第2列について、行列式の線形性から。
3. \begin{vmatrix}
                \overrightarrow{a}_1&\overrightarrow{a}_1
          \end{vmatrix} = \begin{vmatrix}
                                        \overrightarrow{a}_2&\overrightarrow{a}_2
                                      \end{vmatrix} =0,  \begin{vmatrix}
                                                                      \overrightarrow{a}_2&\overrightarrow{a}_1
                                                                     \end{vmatrix} = -\begin{vmatrix}
                \overrightarrow{a}_1&\overrightarrow{a}_2
              \end{vmatrix}