Determinants

2009-12-15 3:41 am
Prove that

│ -bc bc+b^2 bc+c^2 │

│ ca+a^2 -ca ca+c^2 │ = (ab+bc+ca)^3

│ ab+a^2 ab+b^2 -ab │

回答 (1)

2009-12-15 4:25 am
✔ 最佳答案
Row2 = Row2 + Row1 * (ca + a^2)/bc
=> ca+a^2+(-bc)*(ca+a^2)/bc -ca+(bc+b^2)*(ca+a^2)/bc ca+c^2+(bc+c^2)*(ca + a^2)/bc
=> 0 -ca+ca+a^2+ab+a^2b/c ca+c^2+ca+a^2+ac^2/b+a^2c/b
=> 0 (a/c)(ac+bc+ab) (a/b)(ac+ab+bc)+(c/b)(ac+ab+bc)
=> 0 (a/c)(ac+bc+ab) (a/b + c/b)(ac+ab+bc)
Row3 = Row3 + Row1 * (ab + a^2)/bc
=> ab+a^2+(-bc)*(ab+a^2) ab+b^2+(bc+b^2)*(ab+a^2)/bc -ab+(bc+c^2)*(ab+a^2)/bc
=> 0 ab+b^2+ab+a^2+ab^2/c+a^2b/c -ab+ab+a^2+ab^2/c+a^2c/b
=> 0 (a/c+b/c)(ab+ac+bc) (a/b)(ab+bc+ac)
Takes out the factor (ab+ac+bc) from row 2 and 3
(ab + ac + bc)^2 times
| -bc bc+b^2 bc+c^2 |
| 0 a/c (a+c)/b |
| 0 (a+b)/c a/b |
= (ab + ac + bc)^2 * (-bc)[(a/c)(a/b) - (a+c)(a+b)/bc]
= (ab + ac + bc)^2 * (-a^2 + a^2 + ac + ab + bc)
= (ab + ac + bc)^3


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