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Question Number 38960 by Tinkutara last updated on 01/Jul/18

Answered by ajfour last updated on 01/Jul/18

determinant (((xa),(yb),(zc)),((yc),(za),(xb)),((zb),(xc),(ya)))= xa(a^2 yz−bcx^2 ) +by(b^2 zx−acy^2 )+cz(c^2 xy−abz^2 ) = (a^3 +b^3 +c^3 )xyz−abc(x^3 +y^3 +z^3 ) as x+y+z=0 x^3 +y^3 +z^3 = 3xyz , therefore determinant (((xa),(yb),(zc)),((yc),(za),(xb)),((zb),(xc),(ya)))=xyz(a^3 +b^3 +c^3 −3abc) =xyz[a(a^2 −bc)+b(b^2 −ca)+c(c^2 −ab)] = determinant ((a,b,c),(c,a,b),(b,c,a)) .

|xaybzcyczaxbzbxcya|=xa(a2yz−bcx2)+by(b2zx−acy2)+cz(c2xy−abz2)=(a3+b3+c3)xyz−abc(x3+y3+z3)asx+y+z=0x3+y3+z3=3xyz,therefore|xaybzcyczaxbzbxcya|=xyz(a3+b3+c3−3abc)=xyz[a(a2−bc)+b(b2−ca)+c(c2−ab)]=|abccabbca|.

Commented by ajfour last updated on 01/Jul/18

i have no idea how to solve this question, using the properties..

ihavenoideahowtosolvethisquestion,usingtheproperties..

Commented by Tinkutara last updated on 01/Jul/18

Sir why can't we use properties here?

Commented by Tinkutara last updated on 02/Jul/18

OK Sir.

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