​$$\text { If }\left|\begin{array}{ccc}2 x-4 & 4 & 0 \\2 & x-1 & 1 \\2 & 2 & 0\end{array}\right|=0 \text {, then } x=\text {?}$$​​

Correct option is B

​$$\begin{aligned}&\text{Using cofactor expansion along column 3:} \\&\det = 0 \cdot \begin{vmatrix}2 & x - 1 \\2 & 2\end{vmatrix}- 1 \cdot \begin{vmatrix}2x - 4 & 4 \\2 & 2\end{vmatrix}+ 0 \cdot \begin{vmatrix}2x - 4 & 4 \\2 & x - 1\end{vmatrix} \\&\text{Only the \textbf{middle} term survives:} \\&\det = -1 \cdot \begin{vmatrix}2x - 4 & 4 \\2 & 2\end{vmatrix} \\[8pt]&{ \text{ Step 2: Compute the } 2 \times 2 \text{ determinant:}} \\&= -1 \cdot \left[(2x - 4)(2) - (4)(2)\right] = -1 \cdot \left[4x - 8 - 8\right] = -1 \cdot (4x - 16) = -4x + 16 \\[8pt]&{\text{ Step 3: Set determinant to 0}} \\&-4x + 16 = 0 \Rightarrow 4x = 16 \Rightarrow x = 4\end{aligned}$$​​