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Let $$\alpha $$ and $$\beta $$ be the roots of the equation x2 + x + 1 = 0. Then for y $$ \ne $$ 0 in R, $$$\left| {\begin{matrix} {y + 1} & \alpha & \beta \\ \alpha & {y + \beta } & 1 \\ \beta & 1 & {y + \alpha } \\ \end{matrix} } \right|$$$ is equal to

JEE · Math · previous-year question

  1. A.y(y2 – 1)
  2. B.y(y2 – 3)
  3. C.y3correct
  4. D.y3 – 1

Answer

C. y3

Explanation

$$\alpha $$ and $$\beta $$ are the roots of the equation x2 + x + 1 = 0. $$ \therefore $$ $$\alpha $$ = $$\omega $$ and $$\beta $$ = $${\omega ^2}$$ $$\left| {\begin{matrix} {y + 1} & \alpha & \beta \\ \alpha & {y + \beta } & 1 \\ \beta & 1 & {y + \alpha } \\ \end{matrix} } \right|$$ = $$\left| {\begin{matrix} {y + 1} & \omega & {{\omega ^2}} \\ \omega & {y + {\omega ^2}} & 1 \\ {{\omega ^2}} & 1 & {y + \omega } \\ \end{matrix} } \right|$$ C1 $$ \to $$ C1 + C2 + C3 = $$\left| {\begin{matrix} {y + 1 + \omega + {\omega ^2}} & \omega & {{\omega ^2}} \\ {y + 1 + \omega + {\omega ^2}} & {y + {\omega ^2}} & 1 \\ {y + 1 + \omega + {\omega ^2}} & 1 & {y + \omega } \\ \end{matrix} } \right|$$ = $$\left| {\begin{matrix} y & \omega & {{\omega ^2}} \\ y & {y + {\omega ^2}} & 1 \\ y & 1 & {y + \omega } \\ \end{matrix} } \right|$$ As $$1 + \omega + {\omega ^2}$$ = 0 = $$y\left| {\begin{matrix} 1 & \omega & {{\omega ^2}} \\ 1 & {y + {\omega ^2}} & 1 \\ 1 & 1 & {y + \omega } \\ \end{matrix} } \right|$$ R2 $$ \to $$ R2 - R1 R3 $$ \to $$ R3 - R1 = $$y\left| {\begin{matrix} 1 & \omega & {{\omega ^2}} \\ 0 & {y + {\omega ^2} - \omega } & {1 - {\omega ^2}} \\ 0 & {1 - \omega } & {y + \omega - {\omega ^2}} \\ \end{matrix} } \right|$$ = y$$\left[ {\left( {y + {\omega ^2} - \omega } \right)\left( {y + \omega - {\omega ^2}} \right) - \left( {1 - {\omega ^2}} \right)\left( {1 - \omega } \right)} \right]$$ = y(y2) = y3

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