Question

If $\alpha$ and $\beta$ are the roots of the equation $1+x+x^2=0$, then the matrix product
$\left[\begin{array}{cc}1& \beta \\ \alpha & \alpha \end{array}\right]\left[\begin{array}{cc}\alpha & \beta \\ 1& \beta \end{array}\right]$ is equal to

• A. $\left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right]$
• B. $\left[\begin{array}{cc}-1& -1\\ -1& 2\end{array}\right]$
• C. $\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]$
• D. $\left[\begin{array}{cc}-1& -1\\ -1& -2\end{array}\right]$

Answer 1

Answer: (B)

$\left[\begin{array}{cc}-1& -1\\ -1& 2\end{array}\right]$

$\alpha ,\beta$ are roots of $x^2+x+1=0$

$\Rightarrow \alpha =\omega ,\beta ={\omega }^2$

Putting these value in equation given we get,

$\omega +{\omega }^2=-1{\omega }^2+{\omega }^4=-1$

As we know that , $\alpha +\beta =-1$

and $\alpha \beta =1={\omega }^3$

$\left[\begin{array}{cc}1& \beta \\ \alpha & \alpha \end{array}\right]$ $\left[\begin{array}{cc}\alpha & \beta \\ 1& \beta \end{array}\right]=\left[\begin{array}{cc}\alpha +\beta & \beta +{\beta }^2\\ {\alpha }^2+\alpha & \alpha \beta +\alpha \beta \end{array}\right]=\left[\begin{array}{cc}\omega +{\omega }^2& {\omega }^4+{\omega }^2\\ {\omega }^2+\omega & 2{\omega }^3\end{array}\right]$= $\left[\begin{array}{cc}-1& -1\\ -1& 2\end{array}\right]$