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Question: If \(\left| \begin{matrix} \alpha & x & x & x \\ x & \beta & x & x \\ x & x & \gamma & x \\ x & x & ...

If ∣αxxxxβxxxxγxxxxδ∣\left| \begin{matrix} \alpha & x & x & x \\ x & \beta & x & x \\ x & x & \gamma & x \\ x & x & x & \delta \end{matrix} \right| = f(x) – x f '(x), then f(x) is –

A

(x – a) (x – b) (x – g) (x – d)

B

(x + a) (x + b) (x + g) (x + d)

C

2(x – a) (x – b) (x – g) (x – d)

D

None of these

Answer

(x – a) (x – b) (x – g) (x – d)

Explanation

Solution

C2 – C1, C3 – C1, C4 – C1 reduces the given det. In

D=∣αx−αx−αx−αx−(x−β)00x0−(x−γ)0x00−(x−δ)∣\left| \begin{matrix} \alpha & x - \alpha & x - \alpha & x - \alpha \\ x & - (x - \beta) & 0 & 0 \\ x & 0 & - (x - \gamma) & 0 \\ x & 0 & 0 & - (x - \delta) \end{matrix} \right|=a $\left| \begin{matrix}

  • (x - \beta) & 0 & 0 \ 0 & - (x - \gamma) & 0 \ 0 & 0 & - (x - \delta) \end{matrix} \right|$ –x

∣x−αx−αx−α0−(x−γ)000−(x−δ)∣\left| \begin{matrix} x - \alpha & x - \alpha & x - \alpha \\ 0 & - (x - \gamma) & 0 \\ 0 & 0 & - (x - \delta) \end{matrix} \right|

+x $\left| \begin{matrix} x - \alpha & x - \alpha & x - \alpha \

  • (x - \beta) & 0 & 0 \ 0 & 0 & - (x - \delta) \end{matrix} \right|$

– x $\left| \begin{matrix} x - \alpha & x - \alpha & x - \alpha \

  • (x - \beta) & 0 & 0 \ 0 & ( - x - \gamma) & 0 \end{matrix} \right|$

= – a(x – b) (x – g) (x – d) – x (x – a) (x – g)
(x – d) – x (x – a) (x – b) (x – d) – (x – a)
(x – b) (x – g)

= a(x – b) (x – g) (x – d) + x(x – b) (x – g) (x – d) – x(x – b) (x – g) (x – d) – x(x – a) (x – g) (x – d) – x(x – a) (x – b)

(x – d) – x (x – a) (x – b) (x – g) = (x – b) (x – g) (x – d)

(x – a) – x[(x – a) (x – b) (x – g) + (x – b) (x – g) (x – d)

+ (x – g) (x – d) (x – a) + (x – a) (x – b) (x – d)]

= f(x) – x f '(x)

where f(x) = (x – a) (x – b) (x – g) (x – d)