计算行列式Dn+1=,其中ai≠0(i=0,1,2,…,n).

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问题 计算行列式Dn+1,其中ai≠0(i=0,1,2,…,n).

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答案方法一 第i行提出公因子ai (i=2,3,…,n+1),得 [*] =a1a2…an(a0-[*]) 方法二 按最后一列展开,即得 [*] =﹣an-1…a2a1+anDn- =﹣an-1…a2a1+an(﹣an-2…a2a1+an-1Dn-1) =﹣an-1…a2a1-anan-2…a2a1+anan-1Dn-1 =﹣an-1 an-2…a2a1-anan-2…a2a1-…-anan-1…a3a1+anan-1…a3a2D2 =﹣an-1an-2…a2a1-anan-2…a2a1-…-anan-1…a3a1-anan-1…a3a2+anan-1…a1a0

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