Re: [線代] cramer's rule
※ 引述《funnymandui (Dui)》之銘言：
: 請問這題可以幫我解答嗎
: 實在有點不懂
: bcos嘎馬+ccos杯塔=a
: ccos阿法+acos嘎馬=b
: acos杯塔+bcos阿法=c
: apply the cramer's rule to show that
: cos阿法=(b方+c方-a方)/2bc
: 抱歉我不會用符號所以打中文
: 謝謝
是說 這一題 帶公式就弊掉啦
建議原PO翻一下克拉馬法則的通解
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回答一下第一題
本題為一個三角形
且角α、β、γ的對邊長度為a、b、c
令x=cosα, y=cosβ, z=cosγ
原式成為
0x+cy+bz=a
cx+0y+az=b
bx+ay+0z=c
│0 c b│
w=│c 0 a│= 2abc
│b a 0│
│a c b│
│b 0 a│
│c a 0│ ac^2+ab^2-a^3 b^2+c^2-a^2
x=cosα= ──────── = ─────── = ─────── 得證
w 2abc 2bc
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◆ From: 114.37.60.165
※ 編輯: zi6ru04zpgji 來自: 114.37.60.165 (10/30 23:21)
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