If $A=\begin{bmatrix}1 & 3 & 9\\1 & x & x^2\\4& 6 & 9\end{bmatrix}$ is singular, then x is equal to

3 or $\frac{3}{2}$

The correct answer is option (3) → 3 or $\frac{3}{2}$

A → singular ⇒ $|A|=0$

$\begin{bmatrix}1 & 3 & 9\\1 & x & x^2\\4& 6 & 9\end{bmatrix}⇒C_3→C_3-3C_2\begin{bmatrix}1 & 3 & 0\\1 & x & x^2-3x\\4& 6 & -9\end{bmatrix}$

$⇒C_2→C_2-3C_1\begin{bmatrix}1 & 0 & 0\\1 & x-3 & x^2-3x\\4& -6 & -9\end{bmatrix}$

$⇒6x^2-18x-9x+27=0$

$2x^2-6x-3x+9=0$

$2x^2-9x+9=0$

$x=3,\frac{3}{2}$