A) \[x\]
B) \[0\]
C) \[2x\]
D) \[3x\]
Correct Answer: B
Solution :
Given, \[\left| \begin{matrix} x+1 & 2x+1 & 3x+1 \\ 2x & 4x+3 & 6x+3 \\ 4x+4 & 6x+4 & 8x+4 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[2\left| \begin{matrix} 0 & x & 2x \\ 2x & 4x+3 & 6x+3 \\ 2x+2 & 3x+2 & 4x+2 \\ \end{matrix} \right|=0\] Applying \[({{R}_{1}}\to 2{{R}_{1}}-{{R}_{3}})\] \[\Rightarrow \] \[2\left| \begin{matrix} 0 & x & 0 \\ 2x & 4x+3 & -2x-3 \\ 2x+2 & 3x+2 & 4x+2 \\ \end{matrix} \right|=0\] Applying \[({{C}_{3}}\to {{C}_{3}}-2{{C}_{2}})\] \[\Rightarrow \] \[-4\left| \begin{matrix} 0 & x & 0 \\ x & 4x+3 & 2x+3 \\ x+1 & 3x+2 & 2x+2 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[-4x[2{{x}^{2}}+2x-(2x+3)(x+1)]=0\] \[\Rightarrow \] \[-4x[2{{x}^{2}}+2x-(2{{x}^{2}}+5x+3)]=0\] \[\Rightarrow \] \[4x[3x+3]=0\] \[\Rightarrow \] \[x+1=0\] \[(\because \,\,x\ne 0\,\,given)\]
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