Question
Using properties of determinants, solve the following for x:
$ \begin{vmatrix} \text{x - 2} & \text{2x - 3 } & \text{3x - 4 } \\ \text{x - 4} & \text{2x - 9} & \text{2x - 16} \\ \text{x -8} & \text{2x - 27} & \text{3x -64} \end{vmatrix}=0$

Answer

Applying $C_2 \rightarrow C_2 - 2C_1$ and $C_3 \rightarrow C_3 - 3C_1$, we get
$ \begin{vmatrix} \text{x - 2} & \text{1 } & \text{2 } \\ \text{x - 4} & \text{-1} & \text{-4}\\ \text{x -8} & \text{-11} & \text{40} \end{vmatrix}=0$
Applying $R_1 \rightarrow R_1 + R_2$ and $R_3 \rightarrow R_3 - 11R_2$, we get
$ \begin{vmatrix} \text{2x - 6} & \text{0 } & \text{-2 } \\ \text{x - 4} & \text{-1} & \text{-4}\\ \text{-10x + 36} & \text{0} & \text{4} \end{vmatrix}=0$
Expanding along $C_2$, we get –1[8x – 24 – 20x + 72] = 0
12x = 48 i.e. x = 4

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