A) 1, 9
B) -1, 9
C) -1, -9
D) 1, -9
Correct Answer: D
Solution :
By \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\], we have \[(9+x)\] \[\left| \,\begin{matrix} 1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4 \\ \end{matrix}\, \right|\] = 0 \[\Rightarrow \] \[(x+9)\] \[\left| \,\begin{matrix} 0 & 1-x & 0 \\ 0 & -(1-x) & 1-x \\ 1 & 3 & x+4 \\ \end{matrix}\, \right|=0\] \[\Rightarrow \] \[(x+9)\] \[{{(1-x)}^{2}}\left| \,\begin{matrix} 0 & 1 & 0 \\ 0 & -1 & 1 \\ 1 & 3 & x+4 \\ \end{matrix}\, \right|=0\] \[\Rightarrow \] \[x=1,\,1,\,-9\], (Since the determinant = 1).
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