Question

Consider the system of linear equations \[{{x}_{1}}\text{ }+\text{ }2{{x}_{2}}\text{ }+\text{ }{{x}_{3}}\text{ }=\text{ }3\], \[\text{ }2{{x}_{1}}\text{ }+\text{ }3{{x}_{2}}\text{ }+\text{ }{{x}_{3}}\text{ }=\text{ }3\] and \[3{{x}_{1}}\text{ }+\text{ }5{{x}_{2}}\text{ }+\text{ }2{{x}_{3}}\text{ }=\text{ }1\]. The system has
 (a) exactly 3 solutions
 (b) a unique solution
 (c) no solution
 (d) infinite number of solutions

Answer 1

Hint: In order to find the solution of the given question, that is to find the nature of solution of the following \[{{x}_{1}}\text{ }+\text{ }2{{x}_{2}}\text{ }+\text{ }{{x}_{3}}\text{ }=\text{ }3\], \[\text{ }2{{x}_{1}}\text{ }+\text{ }3{{x}_{2}}\text{ }+\text{ }{{x}_{3}}\text{ }=\text{ }3\] and \[3{{x}_{1}}\text{ }+\text{ }5{{x}_{2}}\text{ }+\text{ }2{{x}_{3}}\text{ }=\text{ }1\] system of linear equation by using the following concepts, a system of linear equations has one solution when the graphs intersect at a point; a system of linear equations has no solution when the graphs are parallel or the determinant of the coefficient matrix is equal to zero and a system of linear equations has infinite solutions when the graphs are the exact same line.

Complete step by step solution:
According to the question, given system of equations is as follows:
\[\begin{align}
  & {{x}_{1}}\text{ }+\text{ }2{{x}_{2}}\text{ }+\text{ }{{x}_{3}}\text{ }=\text{ }3 \\
 & 2{{x}_{1}}\text{ }+\text{ }3{{x}_{2}}\text{ }+\text{ }{{x}_{3}}\text{ }=\text{ }3 \\
 & 3{{x}_{1}}\text{ }+\text{ }5{{x}_{2}}\text{ }+\text{ }2{{x}_{3}}\text{ }=\text{ }1 \\
\end{align}\]
In order to find the nature of the solution of the given system of linear equations, we’ll first find the determinant of the coefficient matrix.
The coefficient matrix is \[\left[ \begin{matrix}
   1 & 2 & 1 \\
   2 & 3 & 1 \\
   3 & 5 & 2 \\
\end{matrix} \right]\].
Determinant of the coefficient matrix is as follows:
\[\Rightarrow \left| \begin{matrix}
   1 & 2 & 1 \\
   2 & 3 & 1 \\
   3 & 5 & 2 \\
\end{matrix} \right|\]
Now apply the row matrix operation on the above matrix, that is \[{{R}_{2}}\to {{R}_{2}}+{{R}_{1}}\], we get:
\[\Rightarrow \left| \begin{matrix}
   1 & 2 & 1 \\
   3 & 5 & 2 \\
   3 & 5 & 2 \\
\end{matrix} \right|\]
Apply another row matrix operation on the above matrix, that is \[{{R}_{3}}\to {{R}_{3}}-{{R}_{2}}\], we get:
\[\Rightarrow \left| \begin{matrix}
   1 & 2 & 1 \\
   3 & 5 & 2 \\
   0 & 0 & 0 \\
\end{matrix} \right|=0\]
Clearly, we can see the determinant of the coefficient matrix is zero and we know that a system of linear equations has no solution when the determinant of the coefficient matrix is equal to zero.
Therefore, the given system of linear equations has no solution and hence option (c) is the correct answer.

Note: There’s an alternative way to solve the given question:
A quick observation tells us that the sum of first two equations yields
\[\begin{align}
  & \Rightarrow \left( {{x}_{1}}\text{ }+\text{ }2{{x}_{2}}\text{ }+\text{ }{{x}_{3}} \right)\text{ +}\left( 2{{x}_{1}}\text{ }+\text{ }3{{x}_{2}}\text{ }+\text{ }{{x}_{3}} \right)\text{ }=\text{ }3+3 \\
 & \Rightarrow 3{{x}_{1}}\text{ }+\text{ }5{{x}_{2}}\text{ }+\text{ }2{{x}_{3}}\text{ }=\text{ 6} \\
\end{align}\]
But this contradicts the third equation, i.e.,
\[3{{x}_{1}}\text{ }+\text{ }5{{x}_{2}}\text{ }+\text{ }2{{x}_{3}}\text{ }=\text{ }1\]
As such the system is inconsistent and hence the given system of linear equations has no solution.