Question

lf $\alpha ,\beta$ are the roots of the equation $x^2+x+1=0$ and $S_k={\alpha }^k+{\beta }^k;k=1,2,3,4$ , then
$\left|\begin{array}{ccc}3& 1+S_1& 1+S_2\\ 1+S_1& 1+S_2& 1+S_3\\ 1+S_2& 1+S_3& 1+S_4\end{array}\right|=$

• A. $27$
• B. $-27$
• C. $-3$
• D. $9$

Answer 1

Answer: (B)

$-27$

$S_1=\alpha +\beta =\frac{-b}{a}=-1;\alpha \beta =1$
$S_2={\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =1-2=-1.$
$S_3={\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )=(-1{)}^3-3\left(1\right)(-1)$
$=-1+3=2.$
$S_4={\alpha }^4+{\beta }^4=({\alpha }^2+{\beta }^2)-2{\alpha }^2{\beta }^2=(-1{)}^2-2(1)$
$=-1$
$\left|\begin{array}{ccc}3& 1+S_1& 1+S_2\\ 1+S_1& 1+S_2& 1+S_3\\ 1+S_2& 1+S_3& 1+S_4\end{array}\right|=\left|\begin{array}{ccc}3& 0& 0\\ 0& 0& 3\\ 0& 3& 0\end{array}\right|$
$3(-3\times 3)=-27.$