Answer
Verified
480k+ views
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 $k$.
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 $k$.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE