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Questions  

Let

Δ=0bacaab0cbacbc0, then

equals

a
0
b
abc
c
a2+b2+c2
d
bc+ca+ab

detailed solution

Correct option is A

Interchanging the rows and columns, we get Δ=0a−ba−cb−a0b−cc−ac−b0Taking –1 common from each of R1, R2, R3 we getΔ=(−1)30b−ac−aa−b0c−ba−cb−c0Δ=−Δ⇒2Δ=0 or Δ=0Alternative Solution∆ is a skew symmetric determinant of odd order, therefore ∆ = 0

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