If $\begin{bmatrix} 1 & 3 & 9 \\ 1 & x & x^2 \\ 4 & 6 & 9 \end{bmatrix}$ is singular matrix, where $x \in \mathbb{N}$ (where N set of natural number), then x is equal to

A singular matrix has a determinant equal to zero.

The determinant of the given matrix must equal zero:

$\begin{vmatrix} 1 & 3 & 9 \\ 1 & x & x^2 \\ 4 & 6 & 9 \end{vmatrix} = 0$

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Expanding along the first row:

$\det = 1 \cdot \begin{vmatrix} x & x^2 \\ 6 & 9 \end{vmatrix} - 3 \cdot \begin{vmatrix} 1 & x^2 \\ 4 & 9 \end{vmatrix} + 9 \cdot \begin{vmatrix} 1 & x \\ 4 & 6 \end{vmatrix}$

Computing each 2×2 determinant:

$\begin{vmatrix} x & x^2 \\ 6 & 9 \end{vmatrix} = 9x - 6x^2$

$\begin{vmatrix} 1 & x^2 \\ 4 & 9 \end{vmatrix} = 9 - 4x^2$

$\begin{vmatrix} 1 & x \\ 4 & 6 \end{vmatrix} = 6 - 4x$

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Substituting back:

$\det = 1(9x - 6x^2) - 3(9 - 4x^2) + 9(6 - 4x)$

$= 9x - 6x^2 - 27 + 12x^2 + 54 - 36x$

$= 6x^2 - 27x + 27$

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Setting the determinant equal to zero:

$6x^2 - 27x + 27 = 0$

Dividing by 3:

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

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Factoring the quadratic:

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

$x(2x - 3) - 3(2x - 3) = 0$

$(2x - 3)(x - 3) = 0$

This gives $x = \frac{3}{2}$ or $x = 3$

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Since $x \in \mathbb{N}$ (natural numbers), only $x = 3$ is valid.

Therefore, $x = 3$