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माना $\alpha$ तथा $\beta$ समीकरण $x ^{2}+ x +1=0$ के मूल हैं, तो $R$ में $y \neq 0$ के लिए $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ बराबर है:
$y\,({y^2} - \,3)$
${y^3} - \,1$
$y^3$
$y\,({y^2} - \,1)$
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 ${y^3}$.
