证明$|AB|=|A|·|B| $
标题有误 评论区大佬指出了这个问题，只能说是满足下面条件的矩阵乘法可以有这个性质。
部分条件下 矩阵乘积的行列式=矩阵行列式的乘积
两种情况如下：
- 当A是对角阵的时候， ∣ A B ∣ = ∣ A ∣ ⋅ ∣ B ∣ |AB|=|A|·|B| ∣AB∣=∣A∣⋅∣B∣
证明如下：设 A = { k 1 0 ⋯ 0 0 k 2 ⋯ 0 ⋮ ⋮ ⋯ ⋮ 0 0 ⋯ k n } A=\begin{Bmatrix} k_1&0&\cdots&0 \\0&k_2&\cdots&0\\\vdots&\vdots&\cdots&\vdots\\0&0&\cdots&k_n \end{Bmatrix} A=⎩⎪⎪⎪⎨⎪⎪⎪⎧k10⋮00k2⋮0⋯⋯⋯⋯00⋮kn⎭⎪⎪⎪⎬⎪⎪⎪⎫, B = { b 11 b 12 ⋯ b 1 n b 21 b 22 ⋯ b 2 n ⋮ ⋮ ⋯ ⋮ b n 1 b n 2 ⋯ b n n } B=\begin{Bmatrix} b_{11}&b_{12}&\cdots&b_{1n} \\b_{21}&b_{22}&\cdots&b_{2n}\\\vdots&\vdots&\cdots&\vdots\\b_{n1}&b_{n2}&\cdots&b_{nn} \end{Bmatrix} B=⎩⎪⎪⎪⎨⎪⎪⎪⎧b11b21⋮bn1b12b22⋮bn2⋯⋯⋯⋯b1nb2n⋮bnn⎭⎪⎪⎪⎬⎪⎪⎪⎫
∣ A B ∣ = ∣ k 1 b 11 k 1 b 12 ⋯ k 1 b 1 n k 2 b 21 k 2 b 22 ⋯ k 2 b 2 n ⋮ ⋮ ⋯ ⋮ k n b n 1 k n b n 2 ⋯ k n b n n ∣ = ( ∏ i = 1 n k i ) ⋅ ∣ b 11 b 12 ⋯ b 1 n b 21 b 22 ⋯ b 2 n ⋮ ⋮ ⋯ ⋮ b n 1 b n 2 ⋯ b n n ∣ = ∣ A ∣ ⋅ ∣ B ∣ |AB|=\begin{vmatrix} k_1b_{11}&k_1b_{12}&\cdots&k_1b_{1n} \\k_2b_{21}&k_2b_{22}&\cdots&k_2b_{2n}\\\vdots&\vdots&\cdots&\vdots\\k_nb_{n1}&k_nb_{n2}&\cdots&k_nb_{nn} \end{vmatrix}=\left(\displaystyle\prod_{i=1}^n k_i\right)·\begin{vmatrix} b_{11}&b_{12}&\cdots&b_{1n} \\b_{21}&b_{22}&\cdots&b_{2n}\\\vdots&\vdots&\cdots&\vdots\\b_{n1}&b_{n2}&\cdots&b_{nn} \end{vmatrix}=|A|·|B| ∣AB∣=∣∣∣∣∣∣∣∣∣k1b11k2b21⋮knbn1k1b12k2b22⋮knbn2⋯⋯⋯⋯k1b1nk2b2n⋮knbnn∣∣∣∣∣∣∣∣∣=(i=1∏nki)⋅∣∣∣∣∣∣∣∣∣b11b21⋮bn1b12b22⋮bn2⋯⋯⋯⋯b1nb2n⋮bnn∣∣∣∣∣∣∣∣∣=∣A∣⋅∣B∣
证毕
- 当 A A A是由对角阵 Λ \Lambda Λ经过m次将一列加到另一列上得到时， ∣ A B ∣ = ∣ A ∣ ⋅ ∣ B ∣ |AB|=|A|·|B| ∣AB∣=∣A∣⋅∣B∣
证明如下：设 A = Λ P 1 P 2 . . . P m A=\Lambda P_1P_2…P_m A=ΛP1P2...Pm
∣ A B ∣ = ∣ Λ P 1 P 2 . . . P m B ∣ |AB|=|\Lambda P_1P_2…P_mB| ∣AB∣=∣ΛP1P2...PmB∣
由第一种情况易得：
∣ Λ P 1 P 2 . . . P m B ∣ = ∣ Λ ∣ ⋅ ∣ P 1 P 2 . . . P m B ∣ |\Lambda P_1P_2…P_mB|=|\Lambda|·|P_1P_2…P_mB| ∣ΛP1P2...PmB∣=∣Λ∣⋅∣P1P2...PmB∣
其中 ∣ A ∣ = ∣ Λ ∣ |A|=|\Lambda| ∣A∣=∣Λ∣ ， ∣ P 1 P 2 . . . P m B ∣ = ∣ B ∣ ，|P_1P_2…P_mB|=|B| ，∣P1P2...PmB∣=∣B∣
所以
∣ A B ∣ = ∣ A ∣ ⋅ ∣ B ∣ |AB|=|A|·|B| ∣AB∣=∣A∣⋅∣B∣
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