Consider the determinant \Delta=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{vmatrix} If a_{13}=yz, a_{23}=zx, a_{33}=xy and the minors of a_{13}, a_{23}, a_{33} are respectively (z-y), (z-x), (y-x) then what is the value of \Delta?

  1. A. (z-y)(z-x)(y-x)
  2. B. (x-y)(y-z)(x-z)
  3. C. (x-y)(z-x)(y-z)(x+y+z)
  4. D. (xy+yz+zx)(x+y+z)

Correct Answer: B. (x-y)(y-z)(x-z)

Explanation

Expanding the determinant along the third column yields \Delta = a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33}. Substituting the cofactors C_{ij} = (-1)^{i+j}M_{ij}, we get \Delta = yz(z-y) - zx(z-x) + xy(y-x). Expanding and factoring this polynomial yields (x-y)(y-z)(x-z).

Related questions on Matrices & Determinants

Practice more NDA Mathematics questions