Question

Let $w=\frac{1}{2}+i\frac{\sqrt{3}}{2}$ ,then the value of determinant $\left|\begin{array}{ccc}1& 1& 1\\ 1& -1-w^2& w^2\\ 1& w^2& w^4\end{array}\right|$.

• A. $3w$
• B. $3w(w-1)$
• C. $3w^2$
• D. $3w(1-w)$

Answer 1

Answer: (B)

$3w(w-1)$

$\left|\begin{array}{ccc}1& 1& 1\\ 1& -1-w^2& w^2\\ 1& w^2& w^4\end{array}\right|$. = ?

$=1[-w^4-w^6-w^4]-1[w^4-w^2]+1[w^2+1+w^2]$

$=-2w^4-w^6-w^4+w^2+2w^2+1$

$=-3w^4-w^6+3w^2+1$

$=-3w-1+3w^2+1$ $w^3=1$

$=3w^2-3w$ $w^4=w^3.w=w$

$=3w(w-1)$ $w^6=w^3.w^3=1.1=1$