Question

The value of $\left|\begin{array}{ccc}1+w& w^2& -w\\ 1+w^2& w& -w^2\\ w^2+w& w& -w^2\end{array}\right|$ is equal to

• A. $0$
• B. $2\omega$
• C. $2{\omega }^2$
• D. $-3{\omega }^2$

Answer 1

The correct option is D $-3{\omega }^2$

$\left|\begin{array}{ccc}1+w& w^2& -w\\ 1+w^2& w& {-w}^2\\ w^2+w& w& {-w}^2\end{array}\right|$

Multiplying $C_2$ by $w$ and adding to $C_3$

$C_3\to w{C}_2+C_3$

$\left[\begin{array}{ccc}1+w& w^2& w^2-w\\ 1+w^2& w& 0\\ w^2+w& w& 0\end{array}\right]$

Now finding determinant

$(-w+w^3)\left[w\right(1+w^2)-w(w^2+w\left)\right]$

$(-w+w)(w+w^3-w^3-w^2)$

$\because w^3=1$, then

$(-w+1)(w-w^2)$

$-w^2+w^3+w-w^2$

$-w^2+1+w-w^2$

$-2w^2+1+w+w^2-w^2$

$(1+w+w^2)-3w^2$

As we know , $1+w+w^2=0$

$0-3w^2=-3w^2$

Option $\left(d\right)$ is correct