Question

If $w$ is one cube root of unity, then find the 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. $3w\left(w-1\right)$
• C. $3w^2$
• D. $3w\left(1-w\right)$

Answer 1

The correct option is B

$3w\left(w-1\right)$

Explanation for the correct answer:

Finding the value of the determinant:

Given, $w$ is the cube root of unity

so,

$w^3=1\mathrm{and}{\mathrm{w}}^2+\mathrm{w}+1=0$

From the above two equations, we can modify the given determinant as,

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

Now on solving the above determinant we get,

$\begin{array}{rcl}\left|\begin{array}{ccc}1& 1& 1\\ 1& w& w^2\\ 1& w^2& w\end{array}\right|& =& 1\left(w^2-w^4\right)-1\left(w-w^2\right)+1\left(w^2-w\right)\left[\mathrm{expanding}\mathrm{along}\mathrm{row}1\right]\\ & =& w^2-w-w+w^2+w^2-w\left[\mathbf{\because }\mathbf{}{\mathit{w}}^{\mathbf{4}}\mathbf{=}{\mathit{w}}^{\mathbf{3}}\mathbf{.}\mathit{w}\mathbf{=}\mathit{w}\right]\\ & =& 3w^2-3w\\ & =& 3w\left(w-1\right)\end{array}$

Hence, the correct option is B.