Question

The number of values of k for which the system of linear equations, \[\left( {2k + 1} \right)x + 5ky = k + 2\] and $kx + \left( {k + 2} \right)y = 2$ has no solution, is:
(1) Infinitely many
(2) $3$
(3) $1$
(4) $2$

Answer 1

Hint: A non-homogeneous system of linear equations has a unique solution if its determinant is non-zero. If the determinant is zero, then the system has either no solution. So, for the system to have no solution, the determinant of the system of linear equations must be zero.

Complete step-by-step answer:
Given, system of linear equations:
\[\left( {2k + 1} \right)x + 5ky = k + 2\]
$kx + \left( {k + 2} \right)y = 2$
Above system of linear equation can be written in the form of $AX = B$,
$\left[ {\begin{array}{*{20}{c}}
  {2k + 1}&{5k} \\
  k&{k + 2}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {k + 2} \\
  2
\end{array}} \right]$
Where $A = \left[ {\begin{array}{*{20}{c}}
  {2k + 1}&{5k} \\
  k&{k + 2}
\end{array}} \right],X = \left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
  {k + 2} \\
  2
\end{array}} \right]$
Now, for the system have no solution, determinant of $A$ must be zero.
$\therefore \left| A \right| = 0$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {2k + 1}&{5k} \\
  k&{k + 2}
\end{array}} \right| = 0$
$ \Rightarrow \left( {2k + 1} \right)\left( {k + 2} \right) - 5k \cdot k = 0$
$ \Rightarrow 2{k^2} + 4k + k + 2 - 5{k^2} = 0$
$ \Rightarrow - 3{k^2} + 5k + 2 = 0$
$ \Rightarrow 3{k^2} - 5k - 2 = 0$
$ \Rightarrow 3{k^2} - \left( {6 - 1} \right)k - 2 = 0$
$ \Rightarrow 3{k^2} - 6k + k - 2 = 0$
$ \Rightarrow 3k\left( {k - 2} \right) + 1\left( {k - 2} \right) = 0$
$ \Rightarrow \left( {3k + 1} \right)\left( {k - 2} \right) = 0$
$ \Rightarrow k = \dfrac{{ - 1}}{3}$ or $k = 2$
So, $k$ have $2$ values for which the given system of linear equations has no solution.

Hence, option (4) is the correct answer.

Note: An another method to solve this question is described as follows:
A system of linear equations ${a_1}x + {b_1}y = {c_1}\& {a_2}x + {b_2}y = {c_2}$ has no solution if,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$
Given, system of linear equations:
\[\left( {2k + 1} \right)x + 5ky = k + 2\]
$kx + \left( {k + 2} \right)y = 2$
For no solution:
$\dfrac{{2k + 1}}{k} = \dfrac{{5k}}{{k + 2}} \ne \dfrac{{k + 2}}{2}$
$ \Rightarrow \dfrac{{2k + 1}}{k} = \dfrac{{5k}}{{k + 2}}$
$ \Rightarrow \left( {2k + 1} \right)\left( {k + 2} \right) = 5k \cdot k$
$ \Rightarrow 2{k^2} + 4k + k + 2 = 5{k^2}$
$ \Rightarrow 2{k^2} + 4k + k + 2 = 5{k^2}$
$ \Rightarrow - 3{k^2} + 5k + 2 = 0$
$ \Rightarrow 3{k^2} - 5k - 2 = 0$
$ \Rightarrow 3{k^2} - \left( {6 - 1} \right)k - 2 = 0$
$ \Rightarrow 3{k^2} - 6k + k - 2 = 0$
$ \Rightarrow 3k\left( {k - 2} \right) + 1\left( {k - 2} \right) = 0$
$ \Rightarrow \left( {3k + 1} \right)\left( {k - 2} \right) = 0$
$ \Rightarrow k = \dfrac{{ - 1}}{3}$ or $k = 2$
So, $k$ have $2$ values for which the given system of linear equations has no solution.
Hence, option (4) is the correct answer.