Re: [線代] 矩陣分裂的問題

看板Math作者Annihilator (> No LOVE (%))時間13年前 (2012/12/17 06:54)推噓6(6推 0噓 3→)

留言9則, 3人參與討論串2/2 (看更多)

※ 引述《rnf0 (rnf0)》之銘言： : 大家好: : 請問 : Let A^(2*2) = [2, 3, 1, 2] B^(2*2) = [2, 13, -1, 6] : If AC + CA = B then C = ? : 這種題目除了解 4*4 線性方程組以外，有比較聰明的做法嗎? : 感謝! 令 AY = YD where Y=[y1,y2], D=[λ1,0 ; 0,λ2], Ay1=λ1y1, Ay2=λ2y2 則 By1 = ACy1 + CAy1 = ACy1 + λ1Cy1 = (A+λ1I)Cy1 By2 = ACy2 + CAy2 = ACy2 + λ2Cy2 = (A+λ2I)Cy2 因此 C = Z(Y^-1), where Z=[z1,z2], z1=[(A+λ1I)^-1]By1, z2=[(A+λ2I)^-1]By2 考慮 (A+λ1I)(A+λ2I) = 2(λ1+λ2)A = (2trA)A 得 (2trA)ACy1 = (A+λ1I)(A+λ2I)Cy1 = (A+λ2I)By1 = ABy1 + (detA/λ1)By1 (2trA)ACy2 = (A+λ1I)(A+λ2I)Cy2 = (A+λ1I)By2 = ABy2 + (detA/λ2)By2 合 (2trA)ACY = ABY + (detA)BY(D^-1) (想到更好的) 考慮 A^2 = (trA)A - (detA)I, 則 ABA = A(AC+CA)A = (A^2)CA + AC(A^2) = (trA)ACA-(detA)CA + (trA)ACA-(detA)AC = (2trA)ACA - (detA)B 因 A 可逆且 trA 非零故 1 detA C = ----B + ----(A^-1)B(A^-1) 2trA 2trA = (1/8)[2,13;-1,6] + (1/8)[2,-3;-1,2][2,13;-1,6][2,-3;-1,2] = [1,1;-1,2] -- ～by Jackary P.～ --

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rnf0

12/17 13:16, , 1^F

12/17 13:16, 1^F

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LPH66

12/17 13:43, , 2^F

12/17 13:43, 2^F

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LPH66

12/17 13:44, , 3^F

12/17 13:44, 3^F