LetΔ=0b−ac−aa−b0c−ba−cb−c0, then∆equals

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