A) \[\left| \,\begin{matrix} 2 & 1 & 1 \\ 2 & 2 & 3 \\ 2 & 3 & 6 \\ \end{matrix}\, \right|\]
B) \[\left| \,\begin{matrix} 2 & 1 & 1 \\ 3 & 2 & 3 \\ 4 & 3 & 6 \\ \end{matrix}\, \right|\]
C) \[\left| \begin{matrix} 1 & 2 & 1 \\ 1 & 5 & 3 \\ 1 & 9 & 6 \\ \end{matrix} \right|\]
D) \[\left| \,\begin{matrix} 3 & 1 & 1 \\ 6 & 2 & 3 \\ 10 & 3 & 6 \\ \end{matrix} \right|\,\]
Correct Answer: A
Solution :
\[\left| \,\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \\ \end{matrix}\, \right|\,=\,\left| \,\begin{matrix} 2 & 1 & 1 \\ 3 & 2 & 3 \\ 4 & 3 & 6 \\ \end{matrix}\, \right|\] by \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}\] = \[a,b,c\], by \[{{C}_{2}}\to {{C}_{2}}+{{C}_{3}}\] = \[\left| \,\begin{matrix} 3 & 1 & 1 \\ 6 & 2 & 3 \\ 10 & 3 & 6 \\ \end{matrix}\, \right|\], by \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\]. But \[\ne \left| \,\begin{matrix} 2 & 1 & 1 \\ 2 & 2 & 3 \\ 2 & 3 & 6 \\ \end{matrix}\, \right|\].
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