A) A. P.
B) G. P.
C) H. P.
D) None of these
Correct Answer: B
Solution :
\[\Delta \equiv \left| \,\begin{matrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ a\alpha +b & b\alpha +c & 0 \\ \end{matrix}\, \right|\] = \[\left| \,\begin{matrix} a & b & a\alpha +b \\ b & c & b\alpha +c \\ 0 & 0 & -(a{{\alpha }^{2}}+2b\alpha +c) \\ \end{matrix}\, \right|\], by \[{{R}_{3}}\to {{R}_{3}}-\alpha {{R}_{1}}-{{R}_{2}}\] = \[a\,\{-c(a{{\alpha }^{2}}+2b\alpha +c)-0\}-b\{-b(a{{\alpha }^{2}}+2b\alpha +c)-0\}\] by expanding along \[{{C}_{1}}\] \[=({{b}^{2}}-ac)\,(a{{\alpha }^{2}}+2b\alpha +c)\] Thus, \[\Delta =0\], if either \[{{b}^{2}}-ac=0\]or \[a{{\alpha }^{2}}+2b\alpha +c=0\] i.e., \[a,b,c\] in G.P. or \[a{{\alpha }^{2}}+2b\alpha +c=0\]. Trick: Put \[\alpha =0\], then the determinant \[\left| \,\begin{matrix} a & b & b \\ b & c & c \\ b & c & 0 \\ \end{matrix}\, \right|\,=\,\left| \,\begin{matrix} a & b & 0 \\ b & c & 0 \\ b & c & -c \\ \end{matrix}\, \right|\,=\,-c(ac-{{b}^{2}})=0\]. Hence the result.
You need to login to perform this action.
You will be redirected in
3 sec
