A) \[12\]
B) \[-12\]
C) \[7\]
D) \[-7\]
Correct Answer: B
Solution :
[b] Given, \[\left| \begin{matrix} 2a & 3r & x \\ 4b & 6s & 2y \\ -2c & -3t & -z \\ \end{matrix} \right|=\lambda \left| \begin{matrix} a & r & x \\ b & s & y \\ c & t & z \\ \end{matrix} \right|\] Taking 2 common from \[{{C}_{1}}\] and 3 from \[{{C}_{2}}\] in LHS \[\therefore \,\,\,2\times 3\left| \begin{matrix} a & r & x \\ 2b & 2s & 2y \\ -c & -t & -z \\ \end{matrix} \right|=\lambda \left| \begin{matrix} a & r & x \\ b & s & y \\ c & t & z \\ \end{matrix} \right|\] Taking 2 common from \[{{R}_{2}}\] and \[-1\] from \[{{R}_{3}}\] in LHS \[\therefore -12\left| \begin{matrix} a & r & x \\ b & s & y \\ c & t & z \\ \end{matrix} \right|=\lambda \left| \begin{matrix} a & r & x \\ b & s & y \\ c & t & z \\ \end{matrix} \right|\] \[\Rightarrow \,\lambda =-12\]
You need to login to perform this action.
You will be redirected in
3 sec
