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What will be the value of f(x) if \(\begin{vmatrix}x & b & c\\a & y & c\\a & b & z \end {vmatrix}\)?(a) (x – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} – \frac{c}{z-c}\) – 2)(b) (x – a)(y – b)(z – c)(\(\frac{x}{x-a} – \frac{b}{y – b} – \frac{c}{z-c}\) – 2)(c) (x – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} + \frac{c}{z-c}\) – 2)(d) (x – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} + \frac{c}{z-c}\) + 2)The question was posed to me in exam.Asked question is from Determinant topic in chapter Determinants of Mathematics – Class 12

Answer»

The correct choice is (c) (X – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} + \frac{c}{z-c}\) – 2)

To elaborate: Given, \(\begin{vmatrix}x & b & c\\a & y & c\\a & b & z \END {vmatrix}\) = \(\begin{vmatrix}x & b & c\\a – x & y – b & 0\\0 & b – y & z – c \end {vmatrix}\)

Applying the operation R2 = R2 – R1 and R3 = R3 – R2

= (x – a)(y – b)(z – c)\(\begin{vmatrix}x/(x – a) & b/(y – b) & c/(z – c)\\-1 & y 1 & 0\\0 & -1 & 1 \end {vmatrix}\)

Now, expanding the determinant we get,

= (x – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} + \frac{c}{z-c}\))

= (x – a)(y – b)(z – c)(\(\frac{x}{x-a} + \frac{b}{y – b} + \frac{c}{z-c}\) – 2)

This is because,

\(\frac{b}{y – b} + \frac{c}{z-c} = \frac{y-(y-b)}{y-b} + \frac{z-(z-c)}{z-c} = \frac{y}{y-b}\) – 1 + \(\frac{z}{z-c}\) – 1 = \(\frac{y}{y-b} + \frac{z}{z-c}\) – 2


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