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Property 6

If the determ...

Question

If the determinant

$|\begin{matrix} a & b & 2 a α + 3 b \\ b & c & 2 b α + 3 c \\ 2 a α + 3 b & 2 b α + 3 c & 0 \end{matrix}| = 0, then$

(a) $a, b, c are in H . P$ .
(b) $α is a root of 4 a x^{2} + 12 b x + 9 c = 0 or a, b, c are in G . P .$
(c) $a, b, c are in G . P . only$
(d) $a, b, c are in A . P .$

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Solution

(b) $α is a root of 4 a x^{2} + 12 b x + 9 c = 0 or a, b, c are in G . P .$

$Let Δ = |a b 2 a α + 3 b b c 2 b α + 3 c 2 a α + 3 b 2 b α + 3 c 0| = |a - b b 2 a α + 3 b b - c c 2 b α + 3 c 2 a α + 3 b - 2 b α - 3 c 2 b α + 3 c 0| [Applying C_{1} \to C_{1} - C_{2}] = |a - b b 2 a α + 3 b b - c c 2 b α + 3 c 2 (a - b) α + 3 (b - c) 2 b α + 3 c 0| = |a - b b 2 a α + 3 b b - c c 2 b α + 3 c 0 0 - 2 α (2 a α + 3 b) - 3 (2 b α + 3 c)| [Applying R_{3} \to R_{3} - 2 α, R_{1} - 3 R_{2}] = - 2 α (2 a α + 3 b) - 3 (2 b α + 3 c) |a - b b b - c c| [Expanding along R_{3}] = - (4 a α^{2} + 12 b α + 9 c) (a c - b^{2}) But Δ = 0 [Given] \Rightarrow - (4 a α^{2} + 12 b α + 9 c) (a c - b^{2}) = 0 \Rightarrow (4 a α^{2} + 12 b α + 9 c) = 0 o r (a c - b^{2}) = 0 \Rightarrow α is a root of 4 a x^{2} + 12 b x + 9 c = 0 o r a c = b^{2}, i . e . a, b, c are in G . P .$

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