设a1,a2,a3均为3维列向量，矩阵A=（a1,a2,a3)并且|A|=1，B=（a1+a2+a3,a1+2a2+4a3,a1+3a2+9a3)行列式=？ 设a1,a2,a3均为3维列向量，矩阵A=（a1,a2,a3...
\u8bbea1,a2,a3\u5747\u4e3a3\u7ef4\u5217\u5411\u91cf,\u8bb0\u77e9\u9635A=(a1,a2,a3),B=(a1+a2,a2+a3,an\u884c\uff08\u6216\u8005\u4f60\u5217\uff09\u7684\u884c\u5217\u5f0f\u6ee1\u8db3\uff1a
1. |a1+b1,a2,...,an|=|a1,a2,...,an|+|b1,a2,...,an|;
2. |xa1,a2,...,an|=x|a1,a1,...,an|;
3. |a1,a2,...,a1,...,an|=0\uff08\u5373\u5b58\u5728\u91cd\u590d\u5217\u6216\u884c\uff0c\u5219\u4e3a\u96f6\uff09
\u6839\u636e\u4e0a\u8ff0\u5173\u7cfb\uff0c\u5bb9\u6613\u5f97\u5230\u539f\u77e9\u9635\u7684\u884c\u5217\u5f0f\u662f18+12+9+4+3+2=48
推导一下，对于B的行列式，第三列减去第二列，然后第二列减去第一列，得|a1+a2+a3, a2+3a3, a2+5a3|，然后第三列减去第二列，得|a1+a2+a3, a2+3a3, 2a3|，然后第二列X2，第三列X3，对应行列式前提取所乘数字，得到1/6|a1+a2+a3, 2a2+6a3, 6a3|，然后第二列减去第三列，得1/6|a1+a2+a3, 2a2, 6a3|,提取第二列第三列的乘数，得到2|a1+a2+a3, a2, a3|，然后第一列分别减去第二列和第三列得到 B的行列式最终化简形式 |B|=2|a1, a2, a3|=2|A|=2
绛旓細瑙: (a1+a2+a3, a1+2a2+4a3, a1+3a2+9a3) = (a1,a2,a3)P 鍏朵腑 P = 1 1 1 1 2 3 1 4 9 鍗虫湁 B=AP 鎵浠 |A| = |A||P| = |P| = (2-1)(3-1)(3-2) = 2.娉: |P| 鏄疺andermonde 琛屽垪寮
绛旓細銆愮瓟妗堛戯細D
绛旓細鎴戠殑 璁綼1,a2,a3鍧囦负3缁村垪鍚戦噺,璁扮煩闃礎=(a1,a2,a3)B=(a1+a2+a3,a1+2a 璁綼1,a2,a3鍧囦负3缁村垪鍚戦噺,璁扮煩闃礎=(a1,a2,a3)B=(a1+a2+a3,a1+2a2+2a3,a1+3a2+4a3),濡傛灉|A|=1,閭ｄ箞|B|=... 璁綼1,a2,a3鍧囦负3缁村垪鍚戦噺,璁扮煩闃礎=(a1,a2,a3)B=(a1+a2+a3,a1+2a2+2a3,a1+3a2+4a3),...
绛旓細3. |a1,a2,...,a1,...,an|=0锛堝嵆瀛樺湪閲嶅鍒楁垨琛岋紝鍒欎负闆讹級鏍规嵁涓婅堪鍏崇郴锛屽鏄撳緱鍒板師鐭╅樀鐨勮鍒楀紡鏄18+12+9+4+3+2=48
绛旓細濉 尾2,尾1,尾3 杩欐槸鍥犱负 (伪1,伪2,伪3)P = (尾2,尾1,尾3)
绛旓細A锛a1,a2,a3锛 銆怉(1脳1)锛岋紙a1,a2,a3锛夛紙1脳3锛夛紝绗﹀悎鐭╅樀涔樻硶娉曞垯銆=锛圓*a1,A*a2,A*a3锛=锛圓a1,Aa2,Aa3锛
绛旓細鍙鍥捐鏄庤鍒楀紡绛変簬0锛屾墍浠ョ煩闃典笉鍙嗐傜粡娴庢暟瀛﹀洟闃熷府浣犺В绛旓紝璇峰強鏃堕噰绾炽傝阿璋紒
绛旓細A[(a1+a2) -(a2+a3)]=A(a1 -a3)=0 a1-a3鏄Ax=0鐨勪竴涓В r(A)=2, 鎵浠x=0鐨勮В绌洪棿绉╂槸1锛屾墍浠x=0鐨勮В鏄痥[(a1+a2) -(a2+a3)]鑰孉(a1+a2)=2b => (a1+a2)/2鏄疉x=b鐨勮В 鎵浠x=b 鐨勯氳В鏄 k[(a1+a2) -(a2+a3)] +(a1+a2)/2 ...
绛旓細绠鍗曡绠椾竴涓嬪嵆鍙紝璇︽儏濡傚浘鎵绀
绛旓細4涓3缁鍚戦噺 a1,a2,b1,b2 蹇呯嚎鎬х浉鍏 鎵浠ュ瓨鍦ㄤ竴缁勪笉鍏ㄤ负0鐨勬暟浣垮緱 k1a1+k2a2+k3b1+k4b2 = 0 鍗虫湁 k1a1+k2a2 = -k3b1-k4b2 璁句负 r 鍒檙鏃㈠彲鐢a1a2涔熷彲鐢眀1b2绾挎ц〃鍑, 涓 r 鈮 0 (鍚﹀垯鐢盿1a2绾挎ф棤鍏筹紝b1b2绾挎ф棤鍏,鎺ㄥ嚭 k1=k2=k3=k4=0, 鐭涚浘)
