Show that .

Engineering

Mathematics

Introduction to Determinants

Question

Show that $∣ \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{array} ∣ = (a - b) (b - c) (c - a)$ .

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∣ \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{array} ∣ = (a - b) (b - c) (c - a)

Solving determinant, we have

∣ \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{array} ∣

Applying C₂ → C₂ – C₁ and C₃ → C₃ – C₁

= ∣ \begin{array}{ccc} 1 & 0 & 0 \\ a & b - a & c - a \\ a^{2} & b^{2} - a^{2} & c^{2} - a^{2} \end{array} ∣

Taking (c – a) and (b – a) as common from C₃ and C₂ respectively.

= (b - a) (c - a) ∣ \begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 1 \\ a^{2} & b + a & c + a \end{array} ∣

Expanding the determinant along R₁, we have

= (b – a)(c – a)(1(c + a – b – a) – 0 + 0)

= (b – a) (c – a) (c – b)

= (a – b) (b – c) (c – a)

= R.H.S.

Hence proved.A

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