Question

For what value of $k$ the system of equations kx + 2y = 5 and 3x + y = 1 has no solution?
$
  a.{\text{ }}k = 3 \\
  b.{\text{ }}k = 6 \\
  c.{\text{ }}k \ne 6 \\
  d.{\text{ }}k = 4 \\
$

Answer 1

Hint: Write in matrix form and see the determinant values.

Given system of equations is:
$
  kx + 2y = 5 \\
  3x + y = 1 \\
$
Convert these equations in matrix format
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  k&2 \\
  3&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  5 \\
  1
\end{array}} \right]$
The system of equation has no solution if the value of determinant$\left( D \right) = 0$and at least one of the determinants $\left( {{D_1}{\text{ and }}{D_2}} \right)$is non-zero.
So, ${\text{D = }}\left| {\begin{array}{*{20}{c}}
  k&2 \\
  3&1
\end{array}} \right|$
Now put this determinant to zero and calculate the value of $k$for which system has no solution.
$
   \Rightarrow {\text{D = }}\left| {\begin{array}{*{20}{c}}
  k&2 \\
  3&1
\end{array}} \right| = 0 \\
   \Rightarrow \left( {k \times 1} \right) - \left( {2 \times 3} \right) = 0 \\
   \Rightarrow k = 6 \\
$
So, for$k = 6$, the value of the determinant is zero.
Now, calculate the value of determinant ${D_1}$at this value of$k$.
In determinant ${D_1}$the first column is replaced with column $\left[ {\begin{array}{*{20}{c}}
  5 \\
  1
\end{array}} \right]$
$ \Rightarrow {D_1} = \left| {\begin{array}{*{20}{c}}
  5&2 \\
  1&1
\end{array}} \right| = \left( {5 \times 1} \right) - \left( {2 \times 1} \right) = 3 \ne 0$
Therefore, the system of equations has no solution for $k = 6$
Hence, option b is correct.

Note: - Whenever you face such type of question the key point we have to remember is that put determinant D = 0, then calculate the value of k, then at this value of k if the value of determinant ${D_1}{\text{ or }}{D_2}$ is non-zero then the system of equations has no solution.