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Question 99 of 110

Q.Prove that ∣111xyzx2y2z2∣=(x−y)(y−z)(z−x)\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix} = (x-y)(y-z)(z-x).

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2022Subjective· 3mImportance★★★★★
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Direct expansion along the first row, followed by factoring, shows the determinant equals (x−y)(y−z)(z−x)(x-y)(y-z)(z-x).

Expanding along row 1:

D=1(yz2−zy2)−1(xz2−zx2)+1(xy2−yx2)D = 1(yz^2-zy^2) - 1(xz^2-zx^2) + 1(xy^2-yx^2)

=yz2−y2z−xz2+x2z+xy2−x2y= yz^2-y^2z-xz^2+x^2z+xy^2-x^2y

Group terms with a common factor of x2x^2, then factor progressively:

D=x2(z−y)+x(y2−z2)+yz(z−y)D = x^2(z-y) + x(y^2-z^2) + yz(z-y)

=x2(z−y)−x(z−y)(z+y)+yz(z−y)= x^2(z-y) - x(z-y)(z+y) + yz(z-y) (since y2−z2=−(z−y)(z+y)y^2-z^2=-(z-y)(z+y))

=(z−y)[x2−x(z+y)+yz]= (z-y)\big[x^2 - x(z+y) + yz\big]

=(z−y)[x(x−z)−y(x−z)]= (z-y)\big[x(x-z) - y(x-z)\big] (factoring the quadratic in xx)

=(z−y)(x−z)(x−y)= (z-y)(x-z)(x-y)

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