举个例子,对行列式D:
1+a1 1 1 1 a1 0 0 1
1 1+a2 1 1 →→﹣a2 a2 0 1
1 1 1+a3 1 0 ﹣a3 a3 1
1 1 1 1+a4 0 0 ﹣a4 1+a4
提出 a1a2a3a4 得:
1 0 0 1/a1 1 0 0 1/a1
﹣1 1 0 1/a2 →→ 0 1 0 1/a1+1/a2
0 ﹣1 1 1/a3 0 0 1 1/a1+1/a2+1/a3
0 0 ﹣1 (1+a4)/a4 0 0 0 1/a1+1/a2+1/a3+(1+a4)/a4
=1/a1+1/a2+1/a3+(1+a4)/a4
∴D= a1a2a3a4 ×[1/a1+1/a2+1/a3+(1+a4)/a4 ]
=a1a2a3a4 ×[1+(1/a1)+(1/a2)+(1/a3)+(1/a4)]
同理,得出:所给题目答案为
a1a2a3a4‥‥an×[1+(1/a1)+(1/a2)+(1/a3)+(1/a4)+‥‥‥+(1/an)]