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

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:

& {{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{aligned}$$ In order to find the nature of the solution of the given system of linear equations, we’ll first find the determinant of the coefficient matrix. The coefficient matrix is $$\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{aligned} & \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{aligned}$$ But this contradicts the third equation, i.e., $$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.