周拉的回答：解: Dn = a b b ... b+0 b a b ... b+0 a a b ... a+0 ... ... a a a ... a+(b-a) 按第n列分拆, Dn = a b b ... b b a b ... b a a b ... a ... ... a a a ... a + (b-a)Dn-1 (所有列减第n列, 化为上三角行列式) = (b-a)Dn-1 + a(a-b)^2(b-a)^(n-3) = a(b-a)^(n-1) + (b-a)Dn-1 -- 迭代 = a(b-a)^(n-1) + (b-a)[a(b-a)^(n-2) + (b-a)Dn-2] = 2a(b-a)^(n-1) + (b-a)^2Dn-2 = ... = (n-4)a(b-a)^(n-1) + (b-a)^(n-4)D4 = (n-4)a(b-a)^(n-1) + (b-a)^(n-4)[-(b-a)^4] = [(n-3)a-b](b-a)^(n-1) 注: n>=4 时成立. D3 = -b*(a - b)^2 D4 = -(a - b)^4 D5 = (2*a - b)*(a - b)^4 D6 = -(3*a - b)*(a - b)^5 雄威的回答：解: Dn = a b b ... b+0 b a b ... b+0 a a b ... a+0 ... ... a a a ... a+(b-a) 按第n列分拆, Dn = a b b ... b b a b ... b a a b ... a ... ... a a a ... a + (b-a)Dn-1 (所有列减第n列, 化为上三角行列式) = (b-a)Dn-1 + a(a-b)^2(b-a)^(n-3) = a(b-a)^(n-1) + (b-a)Dn-1 -- 迭代 = a(b-a)^(n-1) + (b-a)[a(b-a)^(n-2) + (b-a)Dn-2] = 2a(b-a)^(n-1) + (b-a)^2Dn-2 = ... = (n-4)a(b-a)^(n-1) + (b-a)^(n-4)D4 = (n-4)a(b-a)^(n-1) + (b-a)^(n-4)[-(b-a)^4] = [(n-3)a-b](b-a)^(n-1) 注: n>=4 时成立. D3 = -b*(a - b)^2 D4 = -(a - b)^4 D5 = (2*a - b)*(a - b)^4 D6 = -(3*a - b)*(a - b)^5 霸王枪的回答：解: d = c1+c2+c3+c4 a+b+c+d b c d a+b+c+d a d c a+b+c+d d a b a+b+c+d c b a r2-r1,r3-r1,r4-r1 a+b+c+d b c d 0 a-b d-c c-d 0 d-b a-c b-d 0 c-b b-c a-d c2+c3 a+b+c+d b+c c d 0 a-b-c+d d-c c-d 0 a-b-c+d a-c b-d 0 0 b-c a-d r3-r2 a+b+c+d b+c c d 0 a-b-c+d d-c c-d 0 0 a-d b-c 0 0 b-c a-d c3+c4 a+b+c+d b+c c d 0 a-b-c+d d-c c-d 0 0 a+b-c-d b-c 0 0 a+b-c-d a-d r4-r3 a+b+c+d b+c c d 0 a-b-c+d d-c c-d 0 0 a+b-c-d b-c 0 0 0 a-b+c-d 行列式 = (a+b+c+d)(a-b-c+d)(a+b-c-d)(a-b+c-d). |
