为什么a=-2时无解：线性方程组增广矩阵(A,b)=(a1,a2,a3,b)，a1=(a,1,1)^T，a2=(1,a,1)^T，a3=(1,1a)^T， 线性代数问题 设四阶可逆矩阵A按列分块为A=(a1，a2，a...
\u8bbe\u5411\u91cfa1=\uff081\uff0c4\uff0c0\uff0c2\uff09\uff0ca2=(2,7\uff0c1\uff0c3\uff09\uff0ca3=(0,1,-1,a),\u03b2=(3,10,b,4)\u6b64\u9898\u4e0d\u80fd\u7528\u7cfb\u6570\u884c\u5217\u5f0f\u6765\u5224\u65ad
\u5e94\u8be5\u7528\u7cfb\u6570\u77e9\u9635\u548c\u5176\u589e\u5e7f\u77e9\u9635\u7684\u79e9
\u5f53r(A)+1=r(A\u589e\u5e7f)\u65f6\uff0c\u65b9\u7a0b\u7ec4\u65e0\u89e3\uff0c\u5373\u03b2\u4e0d\u80fd\u7531a1,a2,a3\u7ebf\u6027\u8868\u51fa
\u6240\u4ee5a\u4e3a\u4efb\u610f\uff0cb\u22602
\u53ea\u6709\u5728\u65b9\u7a0b\u7ec4\u6709\u89e3\u5e76\u4e14\u5176\u589e\u5e7f\u77e9\u9635\u4e3an\u9636\u65b9\u9635\u884c\u5217\u5f0f=0\uff0c\u4e3a\u5145\u8981\u6761\u4ef6
\u53cd\u4e4b\u5219\u4e0d\u4e00\u5b9a\u6210\u7acb,\u5373\u65b9\u7a0b\u7ec4\u65e0\u89e3\u884c\u5217\u5f0f\u22600\uff0c\u4e0d\u80fd\u7528\u8fd9\u4e2a\u6765\u5224\u65ad
\u5b9e\u9645\u4e0a\uff0cA=BP\uff0cB=AP^(-1)
\u5176\u4e2dP\u662f\u521d\u7b49\u5217\u53d8\u6362\u77e9\u9635\uff0c\u662f\u4e24\u4e2a\u521d\u7b49\u77e9\u9635\u7684\u4e58\u79ef\uff1aP=P(1,2)P(1,4)
P^(-1)=P(1,4)P(1,2)
Bx=b\u6709\u552f\u4e00\u89e3x=B^(-1)b=(1\uff0c3\uff0c5\uff0c7)T
\u5219Ax=b\u6709\u552f\u4e00\u89e3x=A^(-1)b=(BP)^(-1)b=P^(-1)B^(-1)b
=P(1,4)P(1,2)B^(-1)b
=(7\uff0c1\uff0c5\uff0c3)T
\u5982\u679c\u4ea4\u6362\u4e00\u6b21\u4f8b\u5982\u65b9\u9635B=(a2\uff0ca1\uff0ca3\uff0ca4)\u7684\u8bdd
\u6b64\u65f6B=P^(-1)
\u5176\u4e2dP\u662f\u521d\u7b49\u5217\u53d8\u6362\u77e9\u9635\uff0cP=P(1,2)
P^(-1)=P(1,2)
Bx=b\u6709\u552f\u4e00\u89e3x=B^(-1)b=(1\uff0c3\uff0c5\uff0c7)T
\u5219Ax=b\u6709\u552f\u4e00\u89e3x=A^(-1)b=(BP)^(-1)b=P^(-1)B^(-1)b
=P(1,2)B^(-1)b
=(3\uff0c1\uff0c5\uff0c7)T
\u7b26\u53f7\u4e0d\u53d8
将每一行加到第三行
第三行为0 0 0 3
那么就使得系数矩阵的秩
小于增广矩阵的秩
于是方程组无解
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绛旓細绯绘暟琛屽垪寮 |A|=a(1-a)鎵浠 a鈮0 涓 a鈮1 鏃舵柟绋缁勬湁鍞竴瑙 褰a=0鏃, 澧炲箍鐭╅樀= 0 1 1 2 1 0 1 2 1 0 1 1 --> 0 1 1 2 1 0 1 2 0 0 0 -1 鏂圭▼缁鏃犺В 褰揳=1鏃, 澧炲箍鐭╅樀= 1 1 1 2 1 1 1 2 1 2 1 1 --> 1...
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