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Question: If ax<sup>3</sup> + bx<sup>2</sup> + cx + d = \(\left| \begin{matrix} x^{2} & (x–1)^{2} & (x–2)^{2}...

If ax3 + bx2 + cx + d

= ∣x2(x–1)2(x–2)2(x–1)2(x–2)2(x–3)2(x–2)2(x–3)2(x–4)2∣\left| \begin{matrix} x^{2} & (x–1)^{2} & (x–2)^{2} \\ (x–1)^{2} & (x–2)^{2} & (x–3)^{2} \\ (x–2)^{2} & (x–3)^{2} & (x–4)^{2} \end{matrix} \right|, Then :

A

a = 1, b = 2, c = 3, d = – 8

B

a = – 1, b = 2, c = 3, d = – 8

C

a = 0, b = 0, c = 0, d = 8

D

a = 0, b = 0, c = 0, d = – 8

Answer

a = 0, b = 0, c = 0, d = – 8

Explanation

Solution

Apply C1 → C1 – C2; C2 → C2 – C3

= ∣(2x–1)(2x–3)(x–2)2(2x–3)(2x–5)(x–3)2(2x–5)(2x–7)(x–4)2∣\left| \begin{matrix} (2x–1) & (2x–3) & (x–2)^{2} \\ (2x–3) & (2x–5) & (x–3)^{2} \\ (2x–5) & (2x–7) & (x–4)^{2} \end{matrix} \right| R1 → R1 – R2 and R2 → R2 – R3

= ∣22(2x–5)22(2x–7)(2x–5)(2x–7)(x–4)2∣\left| \begin{matrix} 2 & 2 & (2x–5) \\ 2 & 2 & (2x–7) \\ (2x–5) & (2x–7) & (x–4)^{2} \end{matrix} \right|

R1 → R1 – R2

= ∣00222(2x–7)(2x–5)(2x–7)(x–4)2∣\left| \begin{matrix} 0 & 0 & 2 \\ 2 & 2 & (2x–7) \\ (2x–5) & (2x–7) & (x–4)^{2} \end{matrix} \right| = – 8

∴Value of determinant is independent of x.

∴ a = b = c = 0 and d = – 8.