Let alpha and beta be the roots of the equa | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$y^3$

Vedclass · Answer

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

$	herefore \Delta  = \left| {\begin{array}{*{20}{c}}
{y + 1}&\omega &{{\omega ^2}}\
\omega &{y + {\omega ^2}}&1\
{{\omega ^2}}&1&{y + \omega }
\end{array}} ight|$

${R_1} 	o {R_1} + {R_2} + {R_3}$

$ \Rightarrow \Delta  = \left| {\begin{array}{*{20}{c}}
1&1&1\
\omega &{y + {\omega ^2}}&1\
{{\omega ^2}}&1&{y + \omega }
\end{array}} ight|$

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 ${y^3}$.

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

Similar Questions

Let $\lambda \in R .$ The system of linear equations

$2 x_{1}-4 x_{2}+\lambda x_{3}=1$

$x_{1}-6 x_{2}+x_{3}=2$

$\lambda x_{1}-10 x_{2}+4 x_{3}=3$ is inconsistent for

In a $\Delta ABC,$ if $\left| {\,\begin{array}{*{20}{c}}1&a&b\\1&c&a\\1&b&c\end{array}\,} \right| = 0$, then ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C = $

$\left| {\,\begin{array}{*{20}{c}}1&5&\pi \\{{{\log }_e}e}&5&{\sqrt 5 }\\{{{\log }_{10}}10}&5&e\end{array}\,} \right| = $

The values of $x $ in the following determinant equation, $\left| {\,\begin{array}{*{20}{c}}{a + x}&{a - x}&{a - x}\\{a - x}&{a + x}&{a - x}\\{a - x}&{a - x}&{a + x}\end{array}\,} \right| = 0$ are

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