3 and 4 .Determinants and Matrices
hard

Let $\alpha $ and $\beta $ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $R,$ $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ is equal to

A

$y\,({y^2} - \,3)$

B

${y^3} - \,1$

C

$y^3$

D

$y\,({y^2} - \,1)$

(JEE MAIN-2019)

Solution

Roots of the equation ${x^2} + x + 1 = 0$ are $\alpha  = \omega $ and $\beta  = {\omega ^2}$ where $\omega ,{\omega ^2}$are complex cube roots of unity

$\therefore \Delta  = \left| {\begin{array}{*{20}{c}}
{y + 1}&\omega &{{\omega ^2}}\\
\omega &{y + {\omega ^2}}&1\\
{{\omega ^2}}&1&{y + \omega }
\end{array}} \right|$

${R_1} \to {R_1} + {R_2} + {R_3}$

$ \Rightarrow \Delta  = \left| {\begin{array}{*{20}{c}}
1&1&1\\
\omega &{y + {\omega ^2}}&1\\
{{\omega ^2}}&1&{y + \omega }
\end{array}} \right|$

Expanding along ${R_1}$, we get

$\Delta  = y.{y^2} \Rightarrow D = {y^3}$

Or

If $\alpha  = {\omega ^2},\beta  = \omega $ we get same value or on expansion using $\alpha  + \beta  =  – 1,\alpha \beta  = 1$ we get value ${y^3}$.

Standard 12
Mathematics

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