If $\begin{vmatrix}a&b&0\\0&a&b\\b&0&a\end{vmatrix}=0$, then
Answer & explanation
Correct answer: option 4
We have,
$\begin{vmatrix}a&b&0\\0&a&b\\b&0&a\end{vmatrix}=0$
$⇒a^3+b^3=0$ [On expanding the determinant on LHS]
$⇒(a/b)^3=-1$
⇒ a/b is one of the cube roots of -1