分块矩阵求行列式，为什么是mXn？ 证明a是mxn矩阵 b是nxm矩阵 n<m时必有行列式ab=...
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\u7531\u5df2\u77e5 AB \u662fmxm\u77e9\u9635
\u7531\u4e8e r(AB)<=r(A)=min{m,n}=n < m
\u6240\u4ee5 AB \u4e3a\u975e\u6ee1\u79e9\u77e9\u9635
\u6545 |AB| = 0.
直接用拉普拉斯定理
详情如图所示
观察矩阵A的分块矩阵，矩阵An和Bm位于副对角线上，要求矩阵行列式|A|，设法通过行列式的行交换的性质把行列式Bm和An换到主对角线上。换行如下，将第一次把Bn的第一行通过和前一行交换，交换n次到第一行，提出来n个-1来
然后是第二行，又是n个-1
总共m行都移到前面去，就是mn个-1
这样就变成一个对角矩阵了，使得|A|=(-1)^nm*|Bn|*|An|。
第一次把Bn的第一行通过和前一行交换，交换n次到第一行，提出来n个-1来
然后是第二行，又是n个-1
总共m行都移到前面去，就是mn个-1
这样就变成一个对角矩阵了
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