线性代数与几何9.计算行列式 ：1.x y ...0 00 x ...0 0" " " "0 0 ...x yy 0 ...0 x (n阶）2.a1-b1 a1-b2 ...a1-bna2-b1 a2-b2 ...a2-bna3-b1 a3-b2 ...a3-bn" " "an-b1 an-b2 ...an-bn

Question

线性代数与几何
9.计算行列式 ：
1.x y ...0 0
0 x ...0 0
' ' ' '
0 0 ...x y
y 0 ...0 x (n阶）
2.a1-b1 a1-b2 ...a1-bn
a2-b1 a2-b2 ...a2-bn
a3-b1 a3-b2 ...a3-bn
' ' '
an-b1 an-b2 ...an-bn

Answer 1

1.解法1:用定义
根据行列式的定义,每行每列恰取一元作乘积
行列式只有两项非零
D = (-1)^t(123...n)x^n + (-1)^t(234...n1)y^n
= x^n +(-1)^(n-1)y^n.
--逆序数 t(234...n1)=n-1
解法2:用展开定理
按第1列展开,行列式 =
x*(-1)^(1+1)*
x y ...0 0
......
0 0 ...x y
0 0 ...0 x
+
y(-1)^(n+1)*
y 0 ...0 0
x y ...0 0
......
0 0 ...x y
= x^n +(-1)^(n+1)y^n.
2.a1-b1 a1-b2 ...a1-bn
a2-b1 a2-b2 ...a2-bn
a3-b1 a3-b2 ...a3-bn
......
an-b1 an-b2 ...an-bn
D1 = a1-b1
D2 = (a1-b1)(a2-b2)-(a1-b2)(a2-b1)
= (a1-a2)(b1-b2)
当n>=3时
Dn =
a1-b1 a1-b2 a1-b3 ...a1-bn
a2-b1 a2-b2 a2-b3 ...a2-bn
a3-b1 a3-b2 a3-b3 ...a3-bn
.........
an-b1 an-b2 an-b3 ...an-bn
c3-c1,c2-c1
a1-b1 b1-b2 b1-b3 ...a1-bn
a2-b1 b1-b2 b1-b3 ...a2-bn
a3-b1 b1-b2 b1-b3 ...a3-bn
.........
an-b1 b1-b2 b1-b3 ...an-bn
第2,3列成比例
所以n>=3时,Dn = 0.