JEE MAIN - Mathematics (2015 (Offline) - No. 11)
The set of all values of $$\lambda $$ for which the system of linear equations:
$$\matrix{ {2{x_1} - 2{x_2} + {x_3} = \lambda {x_1}} \cr {2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}} \cr { - {x_1} + 2{x_2} = \lambda {x_3}} \cr } $$
has a non-trivial solution
$$\matrix{ {2{x_1} - 2{x_2} + {x_3} = \lambda {x_1}} \cr {2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}} \cr { - {x_1} + 2{x_2} = \lambda {x_3}} \cr } $$
has a non-trivial solution
contains two elements
contains more than two elements
in an empty set
is a singleton
Explanation
$$\left. {\matrix{
{2{x_1} - 2{x_2} + {x^3} = \lambda {x_1}} \cr
{2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}} \cr
{\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} = \lambda {x_3}} \cr
} } \right\}$$
$$\eqalign{ & \Rightarrow \,\,\,\,\,\,\,\left( {2 - \lambda } \right){x_1} - 2{x_2} + {x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,2{x_1} - \left( {3 + \lambda } \right){x_2} + 2{x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} - \lambda {x_3} = 0 \cr} $$
For non-trivial solution, $$\Delta = 0$$
i.e. $$\,\,\,\left| {\matrix{ {2 - \lambda } & { - 2} & 1 \cr 2 & { - \left( {3 + \lambda } \right)} & 2 \cr { - 1} & 2 & { - \lambda } \cr } } \right| = 0$$
$$ \Rightarrow \left( {2 - \lambda } \right)\left[ {\lambda \left( {3 + \lambda } \right) - 4} \right] + $$
$$\,\,\,\,\,\,\,\,\,2\left[ { - 2\lambda + 2} \right] + 1\left[ {4 - \left( {3 + \lambda } \right)} \right] = 0$$
$$ \Rightarrow {\lambda ^3} + {\lambda ^2} - 5\lambda + 3 = 0$$
$$ \Rightarrow \lambda = 1,1,3$$
Hence, $$\lambda $$ has $$2$$ values.
$$\eqalign{ & \Rightarrow \,\,\,\,\,\,\,\left( {2 - \lambda } \right){x_1} - 2{x_2} + {x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,2{x_1} - \left( {3 + \lambda } \right){x_2} + 2{x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} - \lambda {x_3} = 0 \cr} $$
For non-trivial solution, $$\Delta = 0$$
i.e. $$\,\,\,\left| {\matrix{ {2 - \lambda } & { - 2} & 1 \cr 2 & { - \left( {3 + \lambda } \right)} & 2 \cr { - 1} & 2 & { - \lambda } \cr } } \right| = 0$$
$$ \Rightarrow \left( {2 - \lambda } \right)\left[ {\lambda \left( {3 + \lambda } \right) - 4} \right] + $$
$$\,\,\,\,\,\,\,\,\,2\left[ { - 2\lambda + 2} \right] + 1\left[ {4 - \left( {3 + \lambda } \right)} \right] = 0$$
$$ \Rightarrow {\lambda ^3} + {\lambda ^2} - 5\lambda + 3 = 0$$
$$ \Rightarrow \lambda = 1,1,3$$
Hence, $$\lambda $$ has $$2$$ values.
Comments (0)
