求解高数行列式证明恒等式|a1+b1x a1x+b1 c1| |a2+b2x a2x+b2 c2||a3+b3x a3x+b3 c3| =(1-x^2)|a1 b1 c1|a2 b2 c2||a2 b2 c2|
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![求解高数行列式证明恒等式|a1+b1x a1x+b1 c1| |a2+b2x a2x+b2 c2||a3+b3x a3x+b3 c3| =(1-x^2)|a1 b1 c1|a2 b2 c2||a2 b2 c2|](/uploads/image/z/8804798-62-8.jpg?t=%E6%B1%82%E8%A7%A3%E9%AB%98%E6%95%B0%E8%A1%8C%E5%88%97%E5%BC%8F%E8%AF%81%E6%98%8E%E6%81%92%E7%AD%89%E5%BC%8F%EF%BD%9Ca1%2Bb1x+a1x%2Bb1+c1%EF%BD%9C+%EF%BD%9Ca2%2Bb2x+a2x%2Bb2+c2%EF%BD%9C%EF%BD%9Ca3%2Bb3x+a3x%2Bb3+c3%EF%BD%9C+%3D%281-x%EF%BC%BE2%29%EF%BD%9Ca1+b1+c1%EF%BD%9Ca2+b2+c2%EF%BD%9C%EF%BD%9Ca2+b2+c2%EF%BD%9C)
求解高数行列式证明恒等式|a1+b1x a1x+b1 c1| |a2+b2x a2x+b2 c2||a3+b3x a3x+b3 c3| =(1-x^2)|a1 b1 c1|a2 b2 c2||a2 b2 c2|
求解高数行列式证明恒等式
|a1+b1x a1x+b1 c1| |a2+b2x a2x+b2 c2||a3+b3x a3x+b3 c3| =(1-x^2)|a1 b1 c1|a2 b2 c2||a2 b2 c2|
求解高数行列式证明恒等式|a1+b1x a1x+b1 c1| |a2+b2x a2x+b2 c2||a3+b3x a3x+b3 c3| =(1-x^2)|a1 b1 c1|a2 b2 c2||a2 b2 c2|
按第1列分拆, 再按第2列分拆, 共分拆为4个行列式的和
(只写第1行)
|a1 a1x c1| + |a1 b1 c1| + |b1x a1x c1| + |b1x b1 c1|
(第1,4个行列式1,2列成比例等于 0)
= |a1 b1 c1| + |b1x a1x c1|
(第2个行列式1,2列提出公因子, 再交换1,2列)
= |a1 b1 c1| - x^2 |a1 b1 c1|
= (1-x^2)|a1 b1 c1|
求解高数行列式证明恒等式:
∣a₁+b₁x a₁x+b₁ c₁∣ ∣a₁ b₁ c₁ ∣
∣a₂+b₂x a₂x+b₂ c₂∣=(1-x²)∣a₂...
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求解高数行列式证明恒等式:
∣a₁+b₁x a₁x+b₁ c₁∣ ∣a₁ b₁ c₁ ∣
∣a₂+b₂x a₂x+b₂ c₂∣=(1-x²)∣a₂ b₂ c₂ ∣
∣a₃+b₃x a₃x+b₃ c₃∣ ∣a₃ b₃ c₃ ∣
证明:【右边第三行写错了吧?按原题二三两行的元素相同会等于零】
∣a₁ a₁x+b₁ c₁∣ ∣b₁x a₁x+b₁ c₁∣
左边=∣a₂ a₂x+b₂ c₂∣+ ∣b₂x a₂x+b₂ c₂∣
∣a₃ a₃x+b₃ c₃∣ ∣b₃x a₃x+b₃ c₃∣
∣a₁ b₁ c₁∣ ∣b₁x a₁x c₁∣
= ∣a₂ b₂ c₂∣+∣b₂x a₂x c₂∣
∣a₃ b₃ c₃ ∣ ∣b₃x a₃x c₃∣
∣a₁ b₁ c₁∣ ∣b₁ a₁ c₁∣
= ∣a₂ b₂ c₂∣+x²∣b₂ a₂ c₂∣
∣a₃ b₃ c₃∣ ∣b₃ a₃ c₃∣
∣a₁ b₁ c₁∣ ∣a₂ b₂ c₂∣
= ∣a₂ b₂ c₂ ∣- x²∣a₃ b₃ c₃∣
∣a₃ b₃ c₃ ∣ ∣a₃ b₃ c₃∣
∣a₁ b₁ c₁∣
= (1-x²)∣a₂ b₂ c₂∣=右边
∣a₃ b₃ c₃∣ ∣a ₁ a₁x c₁ ∣
【其中值为零的行列式没有写出来,如∣a₂ a₂x c₂ ∣=0】
∣a₃ a₃x c₃ ∣
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