⇒Matrices

Correct answer option is (A)

Let $A=\left[\begin{array}{ccc}1& a& b\\ \omega & 1& c\\ {\omega }^2& \omega & 1\end{array}\right]$

For non-singular matrix $|A|\ne 0$

$\Rightarrow \left|\begin{array}{ccc}1& a& b\\ \omega & 1& c\\ {\omega }^2& \omega & 1\end{array}\right|\ne 0$

$\begin{array}{rl}& \Rightarrow 1(1-\omega c)-a(\omega -{\omega }^{2}c)+b(0)\ne 0\\ & \Rightarrow 1(1-\omega c)-a\omega (1-\omega c)\ne 0\\ & \Rightarrow (1-\omega c)(1-a\omega )\ne 0\\ & \Rightarrow c\ne \frac{1}{\omega }\text{ and }a\ne \frac{1}{\omega }\\ & \Rightarrow c\ne {\omega }^2\text{ and }a\ne {\omega }^2\dots [\because {\omega }^3=1]\end{array}$

So possible value of $a$ and $c$ is $\omega$ only and $b$ can take values $\omega$ or ${\omega }^2$.

$\therefore$ The possible number of distinct matrices = 2.

Correct answer option is (A)

$\begin{array}{rl}& \text{ Here }\left|\begin{array}{ccc}1& 2& x\\ 4& -1& 7\\ 2& 4& -6\end{array}\right|=0\\ & \therefore (6-28)-2(-24-14)+x(16+2)=0\\ & \therefore -22+76+18x=0\Rightarrow x=-3\end{array}$