求解这两个行列式的计算结果,需要详细过程,谢谢
1个回答
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1) 提出各行各列的公因子
Dn=(∏ ai)^2*|1+1/a1^2 1 ......... 1|
1 1+1/a2^2 ......... 1
.............................................
1 1 ..... 1+1/an^2
=∏ai^2*|1+1/a1^2 1 ....... 1|
-1/a1^2 1/a2^2 ..... 0 【r2-r1
....................................
-1/a1^2 0 ...... 1/an^2 【rn-r1
=∏ai^2*|∑ 0 ....... 0| 【r1-r2*a2^2-...rn*an^2
-1/a1^2 1/a2^2 ... 0 【∑=(1+a1^2+...+an^2)/a1^2
....................................
-1/a1^2 0 ...... 1/an^2
=∏ai^2*∑*(∏1/ak^2) [ i=1 to n 、k=2 to n ]
=a1^2*∑
=1+a1^2+a2^2+...+an^2
2) r(n-1)+rn/a(n-1)、...、r1+r2/a1
Dn=|∑ 0 0 ...... 0|
m a1 0 ...... 0
n 0 a2 ....... 0
......................
bn-1 ..................... an-1 [∑=a0+b1/a1+b2/a1a2+...+b(n-1)/a1a2..
=∑*∏ai
=a0a1a2..a(n-1)+b1a2a3..a(n-1)+...+b(n-2)a(n-1)+b(n-1)
Dn=(∏ ai)^2*|1+1/a1^2 1 ......... 1|
1 1+1/a2^2 ......... 1
.............................................
1 1 ..... 1+1/an^2
=∏ai^2*|1+1/a1^2 1 ....... 1|
-1/a1^2 1/a2^2 ..... 0 【r2-r1
....................................
-1/a1^2 0 ...... 1/an^2 【rn-r1
=∏ai^2*|∑ 0 ....... 0| 【r1-r2*a2^2-...rn*an^2
-1/a1^2 1/a2^2 ... 0 【∑=(1+a1^2+...+an^2)/a1^2
....................................
-1/a1^2 0 ...... 1/an^2
=∏ai^2*∑*(∏1/ak^2) [ i=1 to n 、k=2 to n ]
=a1^2*∑
=1+a1^2+a2^2+...+an^2
2) r(n-1)+rn/a(n-1)、...、r1+r2/a1
Dn=|∑ 0 0 ...... 0|
m a1 0 ...... 0
n 0 a2 ....... 0
......................
bn-1 ..................... an-1 [∑=a0+b1/a1+b2/a1a2+...+b(n-1)/a1a2..
=∑*∏ai
=a0a1a2..a(n-1)+b1a2a3..a(n-1)+...+b(n-2)a(n-1)+b(n-1)
追问
多谢
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