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正确答案
k(1,2,3,4)T+(1,1,1,1)T,其中k为任意常数
【解析】
根据题设条件知,r(A)=3,所以齐次方程组Ax=0的基础解系含一个非零解.又由α1+2α2+3α3+4α4=0可知,(1,2,3,4)T是Ax=0的一个非零解,由β=α1234可知,(1,1,1,1)T是Ax=β的一个解,因此Ax=β的通解为k(1,2,3,4)T+(1,1,1,1)T,其中k为任意常数.
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