Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

If $ω$ is an imaginary cube root of unity, then the value of $\begin{vmatrix}1+ω&ω^2&-ω\\1+ω^2&ω&-ω^2\\ω^2+ω&ω&-ω^2\end{vmatrix}$ is equal to

Options:

0

$2ω$

$2ω^2$

$-3ω^2$

Correct Answer:

$-3ω^2$

Explanation:

Using $1+ω+ω^2 = 0$, we get

$\begin{vmatrix}1+ω&ω^2&-ω\\1+ω^2&ω&-ω^2\\ω^2+ω&ω&-ω^2\end{vmatrix}$

$=\begin{vmatrix}1+ω+ω^2&ω^2&-ω\\1+ω^2+ω&ω&-ω^2\\ω^2+2ω&ω&-ω^2\end{vmatrix}$  [Applying $C_1→C_1 + C_2$]

$=\begin{vmatrix}0&ω^2&-ω\\0&ω&-ω^2\\ω-1&ω&-ω^2\end{vmatrix}$

$=(ω-1)\begin{vmatrix}ω^2&-ω\\ω&-ω^2\end{vmatrix}$

$=(ω-1)=(-ω^4+ω^2)=(ω-1)=(-ω+ω^2)$

$=-ω^2+ω^3+ω-ω^2=-ω^2+(1+ω)-ω^2=-3ω^2$