If \(\begin{bmatrix}1 & 3 & 9\\1 & x & x^2\\4 & 6 & 9\end{bmatrix}\) is singular matrix, where \(x \in N\) (where N set of natural number), then \(x\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3
A singular matrix has determinant zero. Expanding along the first row:
\( \displaystyle \begin{vmatrix}1 & 3 & 9\\1 & x & x^2\\4 & 6 & 9\end{vmatrix} = 1(9x - 6x^2) - 3(9 - 4x^2) + 9(6 - 4x) \)
\(= 9x - 6x^2 - 27 + 12x^2 + 54 - 36x\)
\(= 6x^2 - 27x + 27\)
Setting it to zero:
\(6x^2 - 27x + 27 = 0\)
\(2x^2 - 9x + 9 = 0\)
\((x - 3)(2x - 3) = 0\)
\( \displaystyle x = 3 \text{ or } x = \frac{3}{2} \)
Since x is a natural number, \(x = 3\).
Shortcut: For \(x = 3\) the second row becomes (1, 3, 9), the same as the first row, and a determinant with two identical rows is zero.