A) \[x=0\]
B) \[x=6\]
C) \[x=3\]
D) None of these
Correct Answer: A
Solution :
Given that, \[\left| \begin{matrix} 3-x & -6 & 3 \\ -6 & 3-x & 3 \\ 3 & 3 & -6-x \\ \end{matrix} \right|=0\] Applying \[({{R}_{1}}\to {{R}_{1}}-{{R}_{2}})\] \[\Rightarrow \] \[\left| \begin{matrix} 9-x & -9+x & 0 \\ -6 & 3-x & 3 \\ 3 & 3 & -6-x \\ \end{matrix} \right|=0\] Taking out \[(9-x)\] common from \[{{R}_{1}}\] \[\Rightarrow \] \[(9-x)\left| \begin{matrix} 1 & -1 & 0 \\ -6 & 3-x & 3 \\ 3 & 3 & -6-x \\ \end{matrix} \right|=0\] Applying \[({{C}_{1}}\to {{C}_{1}}+{{C}_{2}})\] \[\Rightarrow \] \[(9-x)\left| \begin{matrix} 0 & -1 & 0 \\ -3-x & 3-x & 3 \\ 6 & 3 & -6-x \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[(9-x)[1\{(-6-x)(-3-x)-3\times 6\}]=0\] \[\Rightarrow \] \[(9-x)(18+6x+3x+{{x}^{2}}-18)=0\] \[\Rightarrow \] \[(9-x)({{x}^{2}}+9x)=0\] \[\Rightarrow \] \[(9-x)x(x+9)=0\] \[\Rightarrow \] \[x=0,\,\,9,-9\]
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