The determinant Δ=b2−ab b−c bc−acab−a2 a−b b2−abbc−ac c−a ab−a2 equals
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a
(b – c) (c – a) (a – b)
b
abc (b – c) (c – a) (a – b)
c
(a + b + c) (b – c) (c – a) (a – b)
d
0
answer is D.
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Detailed Solution
Write Δ=Δ1−Δ2 whereΔ1=b2−abbbc−acab−a2ab2−abbc−accab−a2 and Δ2=b2−abcbc−acab−a2bb2−abbc−acaab−a2In Δ1 use C1→C1−(b−a)C2 to show that Δ1=0.In Δ2 use C3→C3−(b−a)C2 to show that ∆2=0.Thus, ∆=0.