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.