一元三次求行列式
若A、B、R為方程式X^3-6X^2+10X-7=0 三根，
則行列式A B R﹀B R A﹀R A B =
(1)X^3-6X^2+10X-7=0
由根與係數得: A+B+R=6, AB+BR+RA=10, ABR=7
A^3+B^3+R^3-3ABR= (A+B+R)(A^2+B^2+R^2-AB-BR-RA)
=6*[(A+B+R)^2-3(AB+BR+RA)]=6*(36-30)=36
行列式A B R﹀B R A﹀R A B
=3ABR-(A^3+B^3+R^3)
=-(A^3+B^3+R^3-3ABR)
=-36
由根與係數得: A+B+R=6, AB+BR+RA=10, ABR=7
A^3+B^3+R^3-3ABR= (A+B+R)(A^2+B^2+R^2-AB-BR-RA)
=6*[(A+B+R)^2-3(AB+BR+RA)]=6*(36-30)=36
行列式A B R﹀B R A﹀R A B
=3ABR-(A^3+B^3+R^3)
=-(A^3+B^3+R^3-3ABR)
=-36
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