will be non-singular, are
A) \[-2\le x\le 2\]
B) for all\[x\]other than 2 and\[-2\]
C) \[x\ge 2\]
D) \[x\le -2\]
Correct Answer: B
Solution :
Let\[\left[ \begin{matrix} -x & x & 2 \\ 2 & x & -x \\ x & -2 & -x \\ \end{matrix} \right]\]is singular. \[\therefore \] \[\left| \begin{matrix} -x & x & 2 \\ 2 & x & -x \\ x & -2 & -x \\ \end{matrix} \right|=0\] Applying \[{{C}_{2}}\to {{C}_{2}}+{{C}_{1}},{{C}_{3}}\to {{C}_{3}}+{{C}_{1}}\] \[\Rightarrow \] \[\left| \begin{matrix} -x & 0 & 2-x \\ 2 & 2+x & 2-x \\ x & x-2 & 0 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[(2-x)\left| \begin{matrix} -x & 0 & 1 \\ 2 & 2+x & 1 \\ x & x-2 & 0 \\ \end{matrix} \right|=0\] Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] \[\Rightarrow \] \[(2-x)\left| \begin{matrix} -x & 0 & 1 \\ 2+x & 2+x & 0 \\ x & x-2 & 0 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[(2-x)(2+x)\left| \begin{matrix} -x & 0 & 1 \\ 1 & 1 & 0 \\ x & x-2 & 0 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[(2-x)(2+x)(x-2-x)=0\] \[\Rightarrow \] \[-2(2-x)(2+x)=0\] \[\Rightarrow \] \[x=-2,2\] \[\therefore \]Given matrix is non-singular for all\[x\]other than 2 and\[-2\].
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