线性代数行列式证明题！

Question

证明： 1 a^2 a^3 1 a a^2 1 b^2 b^3 = (ab+bc+ca) 1 b b^2 1 c^2 c^3 1 c c^2

Answer 1

^* XA=2XA-8E,A^* XA-2XA=-8E,A的行列式为det(A)=-2,A^-1=A^*det(A),故 -2A^* XA+4XA=16E A^-1 XA+4XA=16E (A^-1 +4E)XA=16E XA=16(A^-1 +4E)^-1 X=16(A^-1 +4E)^-1A^-1=16(A(A^-1 +4E))^-1=16((E +4A))^-1 E +4A= 5 ,0 ,0 0 ,-7 ,0 0 ,0 ,5 对角阵的逆等于对角线元求倒数 16((E +4A))^-1 16/5 ,0 ,0 0 ,-16/7 ,0 0 ,0 ,16/5 故X= 16/5 ,0 ,0 0 ,-16/7 ,0 0 ,0 ,16/5 这个问题线性代数行列式证明题！,好难啊，辛辛苦苦回答了，给我个满意答案把

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