A) a and b
B) b and c
C) a and c
D) All of these
Correct Answer: B
Solution :
[b] \[Let\,\,D=\left| \begin{matrix} a+b+c & a+b & a \\ 4a+3b+2c & 3a+2b & 2a \\ 10a+6b+3c & 6a+3b & 3a \\ \end{matrix} \right|\] \[\Rightarrow \,\,D=\left| \begin{matrix} a+b+c & a+b & a \\ 4a+3b+2c & 3a+2b & 2a \\ 10a+6b+3c & 6a+3b & 3a \\ \end{matrix} \right|\] By \[{{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}\] and \[{{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}\], we get: \[\Rightarrow \left| \begin{matrix} a+b+c & a+b & a \\ 2a+b & a & 0 \\ 7a+3b & 3a & 0 \\ \end{matrix} \right|\] By \[{{C}_{1}}\to {{C}_{1}}-{{C}_{2}}\] gives: \[\Rightarrow \left| \begin{matrix} c & a+b & a \\ a+b & a & 0 \\ 4a+3b & 3a & 0 \\ \end{matrix} \right|\] Again by \[{{R}_{3}}\to {{R}_{3}}-3{{R}_{1}},\] we get: \[D=\left| \begin{matrix} a+b+c & a+b & 0 \\ a+b & a & 0 \\ a & 0 & 0 \\ \end{matrix} \right|\] \[=a\{0.(a+b)-a.a\}\] \[=-{{a}^{3}}\] which is independent of b and c.
You need to login to perform this action.
You will be redirected in
3 sec
