Q.If $\begin{vmatrix} 2x & 5 \\ 4 & x \end{vmatrix} = \begin{vmatrix} 3 & 5 \\ 4 & 6 \end{vmatrix}$, then the value of $x$ is (A) $\dfrac{3}{2}$ (B) $6$ (C) $3$ (D) $\pm 3$

Expanding each $2\times2$ determinant (product of the main diagonal minus product of the off-diagonal) turns the equation into $2x^2-20=-2$. Sol …

1. Expand the left determinant: $\begin{vmatrix}2x&5\\4&x\end{vmatrix}=2x\cdot x - 5\cdot 4 = 2x^2-20$.
2. Expand the right determinant: $\begin{vmatrix}3&5\\4&6\end{vmatrix}=3\cdot6-5\cdot4=18-20=-2$. …