A) 3,-1
B) -3, 1
C) 3, 1
D) -3,-1
Correct Answer: A
Solution :
Since \[\left| \begin{matrix} x & 2 & -1 \\ 2 & 5 & x \\ -1 & 2 & x \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} x & 2 & -1 \\ 2 & 5 & x \\ -3 & -3 & 0 \\ \end{matrix} \right|=0\] \[{{R}_{3}}\to {{R}_{3}}-{{R}_{2}}\] \[\Rightarrow -1\,(-6+15)-x\,[-3x+6]=0\] \[\Rightarrow \] \[-9+3{{x}^{2}}-6x=0\] \[\Rightarrow \] \[{{x}^{2}}-2x-3=0\] \[\Rightarrow \] \[(x-3)\,(x+1)=0\] \[\Rightarrow \] \[x=-1,3\]
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