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JEE Main 2021 · Mathematics · Matrices and Determinants · medium

If a,b,c are the roots of t^3-6t^2+11t-6=0, then beginvmatrixa&b&c\b&c&a\c&a&bendvmatrix equals

Single correct option

If a,b,ca,b,c are the roots of t3−6t2+11t−6=0t^3-6t^2+11t-6=0, then ∣abcbcacab∣\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix} equals
  1. A. 1818
  2. B. −6-6
  3. C. 00
  4. D. −18-18
Show answer and solution

Answer: (D) −18-18

Roots are 1,2,31,2,3 since t3−6t2+11t−6=(t−1)(t−2)(t−3)t^3-6t^2+11t-6=(t-1)(t-2)(t-3). For the circulant-type determinant, ∣abcbcacab∣=3abc−a3−b3−c3=−(a+b+c)(a2+b2+c2−ab−bc−ca).\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}=3abc-a^3-b^3-c^3=-(a+b+c)\big(a^2+b^2+c^2-ab-bc-ca\big). Here a+b+c=6a+b+c=6 and a2+b2+c2−ab−bc−ca=14−11=3a^2+b^2+c^2-ab-bc-ca=14-11=3. So the determinant is −6⋅3=−18-6\cdot3=-18. Direct expansion with (a,b,c)=(1,2,3)(a,b,c)=(1,2,3) also gives −18-18. The factorisation of the cubic is used only to read off a+b+c=6a+b+c=6 and ab+bc+ca=11ab+bc+ca=11.
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