Question

Determine the value of $k$ so that following linear equations have no solutions:
$\left( {3k + 1} \right)x + 3y - 2 = 0 \\\ \left( {{k^2} + 1} \right)x + \left( {k - 2} \right)y - 5 = 0 \\\$

Answer 1

Hint: Convert the equations into matrix format and Equate the determinant $D$ of this matrix to zero using this to calculate the value of k to reach the answer.

Complete step-by-step answer:
Given system of equations is:
$\left( {3k + 1} \right)x + 3y - 2 = 0 \\\ \left( {{k^2} + 1} \right)x + \left( {k - 2} \right)y - 5 = 0 \\\$
Convert these equations into matrix format
\Rightarrow \left[ {\begin{array}{*{20}{c}} {\left( {3k + 1} \right)}&3 \\\ {\left( {{k^2} + 1} \right)}&{\left( {k - 2} \right)} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \\\ 5 \end{array}} \right]
The system of equations has no solutions if the value of determinant $\left( D \right) = 0$, and at least one of the determinant $\left( {{D_1}{\text{ and }}{{\text{D}}_2}} \right)$ is non-zero.
So, determinant $\left( D \right)$ of the above system of equations is given below
So, {\text{D = }}\left| {\begin{array}{*{20}{c}} {\left( {3k + 1} \right)}&3 \\\ {\left( {{k^2} + 1} \right)}&{\left( {k - 2} \right)} \end{array}} \right|
Now put this determinant equal to zero and calculate the value of $k$ for which the system of equations has no solution.
\Rightarrow {\text{D = }}\left| {\begin{array}{*{20}{c}} {\left( {3k + 1} \right)}&3 \\\ {\left( {{k^2} + 1} \right)}&{\left( {k - 2} \right)} \end{array}} \right| = 0 \\\ \Rightarrow \left( {3k + 1} \right)\left( {k - 2} \right) - \left( {{k^2} + 1} \right)3 = 0 \\\ \Rightarrow 3{k^2} - 6k + k - 2 - 3{k^2} - 3 = 0 \\\ \Rightarrow - 5k - 5 = 0 \\\ \Rightarrow k = - 1 \\\
So, for $k = - 1$, the value of determinant is zero.
Now calculate the value of determinant ${{\text{D}}_1}$ at this value of $k$, to ensure that the condition is satisfied for no solution. The value of determinant ${{\text{D}}_1}$ should not be equal to zero.
If first column is replaced with column \left[ {\begin{array}{*{20}{c}} 2 \\\ 5 \end{array}} \right]{\text{ }}, then the determinant D is converted into determinant ${{\text{D}}_1}$, according to Cramer Rule.
\Rightarrow {{\text{D}}_1} = \left| {\begin{array}{*{20}{c}} 2&3 \\\ 5&{\left( {k - 2} \right)} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 2&3 \\\ 5&{\left( { - 1 - 2} \right)} \end{array}} \right| = 2\left( { - 1 - 2} \right) - 3 \times 5 = - 6 - 15 = - 21 \ne 0
Therefore the system of equations has no solution for $k = - 1$.

Note: Whenever we face such types of questions the key concept we have to remember is that put determinant ${\text{D = 0}}$, then calculate the value of $k$, then at this value of $k$ if the value of determinant ${{\text{D}}_1}{\text{ or }}{{\text{D}}_2}$ is non-zero then the system of equations has no solution at this value of $k$.