InterviewSolution
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InterviewSolution
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Without expanding, show that the value of each of the following determinants is zero:\(\begin{vmatrix}1 &a &a^2-bc \\[0.3em]1 & b & b^2-ac \\[0.3em]1 &c & c^2-ab\end{vmatrix}\) |
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Answer» Let Δ = \(\begin{vmatrix}1 &a &a^2-bc \\[0.3em]1 & b & b^2-ac \\[0.3em]1 &c & c^2-ab\end{vmatrix}\) ⇒ Δ = \(\begin{vmatrix}1 &a &a^2 \\[0.3em]1 & b & b^2\\[0.3em]1 &c & c^2\end{vmatrix}\) - \(\begin{vmatrix}1 &a &bc \\[0.3em]1 & b & ac \\[0.3em]1 &c & ab\end{vmatrix}\) Applying R2→R2 – R1 and R3 → R3 – R1, we get, ⇒ Δ = \(\begin{vmatrix}1 &a &a^2 \\[0.3em]0 & b-a & b^2-a^2\\[0.3em]0 &c-a & c^2-a^2\end{vmatrix}\) - \(\begin{vmatrix}1 &a &bc \\[0.3em]0 & b-a & (a-b)c \\[0.3em]0 &c-a & (a-c)b\end{vmatrix}\) Taking (b – a) and (c – a) common from R2 and R3 respectively, ⇒ Δ = (b-a)(c-a) \(\begin{vmatrix}1 &a &a^2 \\[0.3em]0 & 1 & b+a\\[0.3em]0 &1 & c+a\end{vmatrix}\) - (b-a)(c-a) \(\begin{vmatrix}1 &a &bc \\[0.3em]0 & 1 & -c \\[0.3em]0 &1 & -b\end{vmatrix}\) = [(b – a)(c – a)][(c + a) – (b + a) – ( – b + c)] = [(b – a)(c – a)][c + a + b – a – b – c] = [(b – a)(c – a)][0] = 0 |
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