Question

If $\begin{vmatrix}x&x+y&x+y+z\\ 2x&3x+2y&4x+3y+2z\\ 3x&6x+3y&10x+6y+3z\end{vmatrix}= 64 $ then the real value of x is :

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (C)

Let $\Delta = \begin{vmatrix}x&x+y&x+y+z\\ 2x&3x+2y&4x+3y+2z\\ 3x&6x+3y&10x+6y+3z\end{vmatrix}$
$= \begin{vmatrix}x&x+y&x+y+z\\ 0&x&2x+y\\ 0&3x&7x+3y\end{vmatrix} $
using $R_{2} \to R_{2} - 2R_{1}R_{3} \to R_{3} - 3R_{1}$
$ = \begin{vmatrix}x&x+y&x+y+z\\ 0&x&2x+y\\ 0&0&x\end{vmatrix}$ using $R_{3} \to R_{3} - 3R_{2} = x^3 $
Giver $\Delta= 64 $
$\therefore x^{3} = 64 \Rightarrow x=4 $