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If f(x) $= (x - \alpha)(x - \beta)(x - \gamma)(x - \delta),$ then the determinant

$$ \begin{vmatrix} \alpha & x & x & x \ x & \beta & x & x \ x & x & \gamma & x \ x & x & x & \delta \end{vmatrix} $$ is equal to

A. f'(x) B. xf'(x) C. f(x) + xf'(x) D. f(x) - xf'(x)


University Name > Engineering > Sem > Mathematics

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SOLUTION :

$$ \begin{vmatrix} \alpha & x & x & x \\ x & \beta & x & x \\ x & x & \gamma & x \\ x & x & x & \delta \end{vmatrix} $$ $$ = \begin{vmatrix} \alpha & x-\alpha & x-\alpha & x-\alpha \\ x & \beta - x & 0 & 0 \\ x & 0 & \gamma - x & 0 \\ x & 0 & 0 & \delta - x \end{vmatrix} $$

$= \alpha (\beta - x)(\gamma - x)(\delta - x) - x[(x - \alpha)(x - \gamma)(x - \delta) + (x - \alpha)(x - \beta)(x - \delta) + (x - \alpha)(x - \beta)(x - \gamma)]$

$= (x - \alpha)(x - \beta)(x - \gamma)(x - \delta) - x[(x - \alpha)(x - \beta)(x - \gamma) + (x - \alpha)(x - \beta)(x - \delta) + (x - \alpha)(x - \gamma)(x - \delta) + (x - \beta)(x - \gamma)(x - \delta)]$

$= f(x) - xf'(x) $

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