If \( \begin{vmatrix}x-2 & 2x-3 & 3x-4\\x-4 & 2x-9 & 3x-16\\x-8 & 2x-27 & 3x-64\end{vmatrix}=0 \) then the value of \( x \) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4
With the row operations \( R_2\to R_2-R_1 \) and \( R_3\to R_3-R_1 \), the x terms cancel:
\( \displaystyle \begin{vmatrix}x-2 & 2x-3 & 3x-4\\-2 & -6 & -12\\-6 & -24 & -60\end{vmatrix}=0 \)
Taking −2 common from \( R_2 \) and −6 from \( R_3 \):
\( \displaystyle 12\begin{vmatrix}x-2 & 2x-3 & 3x-4\\1 & 3 & 6\\1 & 4 & 10\end{vmatrix}=0 \)
Expanding along \( R_1 \):
\( \displaystyle (x-2)(30-24)-(2x-3)(10-6)+(3x-4)(4-3)=0 \)
\( \displaystyle 6x-12-8x+12+3x-4=0 \)
\( \displaystyle x-4=0 \)
\( \displaystyle x=4 \)