线性代数~第三题求解答,需要过程
\u7ebf\u6027\u4ee3\u6570\u9898\uff0c\u7b2c\u4e09\u9898\uff0c\u6c42\u8fc7\u7a0b\u6c42\u89e3\uff0c\u5229\u7528\u4e24\u4e2a\u6392\u5217\u7684\u9006\u5e8f\u6570\u4e4b\u548c\uff1dC(n\uff0c2)
\u6240\u6c42\u6392\u5217\u7684\u9006\u5e8f\u6570\uff1dC(n\uff0c2)\uff0ds
\u8fc7\u7a0b\u5982\u4e0b\u56fe\uff1a
A*A-1=E
A\u884c\u5217\u5f0f*A\u9006\u884c\u5217\u5f0f=1
\u7ed9\u4f60\u7167\u7247\u89e3\u8bf4\u3002\u3002
|A3,5A2,A1|
而提取去每行的5,
即5|A3,A2,A1|
那么|A3,A2,A1|与|A|相比就交换了1行,
所以需要乘以 -1
于是得到|A3,A2,A1|= -|A|=3
所以所求的行列式就等于
5|A3,A2,A1|= -5|A|= 15
绛旓細鍗5|A3,A2,A1| 閭d箞|A3,A2,A1|涓巪A|鐩告瘮灏变氦鎹簡1琛岋紝鎵浠ラ渶瑕佷箻浠 -1 浜庢槸寰楀埌|A3,A2,A1|= -|A|=3 鎵浠ユ墍姹傜殑琛屽垪寮忓氨绛変簬 5|A3,A2,A1|= -5|A|= 15
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