Question

Consider the system of linear equations:
$\begin{aligned} & {{x}_{1}}+2{{x}_{2}}+{{x}_{3}}=3, \\\ & 2{{x}_{1}}+3{{x}_{2}}+{{x}_{3}}=3, \\\ & 3{{x}_{1}}+5{{x}_{2}}+2{{x}_{3}}=1 \\\ \end{aligned}$
The system has:
(a) Infinite number of solutions
(b) Exactly 3 solutions
(c) A unique solution
(d) No solution

Answer 1

Hint: First, we should know the formula to calculate the determinant of the matrix expanding it with row 1 of the matrix as $\det ={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)-{{a}_{12}}\left( {{a}_{21}}{{a}_{33}}-{{a}_{23}}{{a}_{31}} \right)+{{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right)$. Then, we have to find the determinant of A as |A| to get the system of linear equations. Then, on getting the determinant of A we can conclude what type of system of linear equations they are.

Complete step by step solution:
In this question, we are supposed to find the behaviour of the linear equations by using the relation that if the determinant of the coefficients of the linear variables from the given question is calculated it tells us about the behaviour.
Let denote the coefficients matrix as A. Find the matrix A from the given question as:
$A=\left( \begin{matrix} 1 & 2 & 1 \\\ 2 & 3 & 1 \\\ 3 & 5 & 2 \\\ \end{matrix} \right)$
Now by using the condition of determinant that if |A| is zero then, the system of linear equations has no solution.
By finding the determinant of matrix as |A| as:

1 & 2 & 1 \\\ 2 & 3 & 1 \\\ 3 & 5 & 2 \\\ \end{matrix} \right|$$ Now, for finding the determinant we need to expand it along the row ${{R}_{1}}$ which is the first row of the matrix. Use the formula for finding the matrix as: $\det ={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)-{{a}_{12}}\left( {{a}_{21}}{{a}_{33}}-{{a}_{23}}{{a}_{31}} \right)+{{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right)$ The above formula has coefficients named by following representation as: $\left( \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right)$ Now, by finding the determinant of |A|, we can comment on the system of linear equations as: $$\begin{aligned} & \left| A \right|=1\left( 3\times 2-1\times 5 \right)-2\left( 2\times 2-1\times 3 \right)+1\left( 2\times 5-3\times 3 \right) \\\ & \Rightarrow 1\left( 6-5 \right)-2\left( 4-3 \right)+1\left( 10-9 \right) \\\ & \Rightarrow 1\left( 1 \right)-2\left( 1 \right)+1\left( 1 \right) \\\ & \Rightarrow 1-2+1 \\\ & \Rightarrow 0 \\\ \end{aligned}$$ The above determinant proves that the system of linear equations has no solution as if determinant |A| is zero then, the system of linear equations has no solution. Hence, option (d) is the correct answer. Note: Another approach to solve this kind of problem is as given below: A quick observation tells us that the sum of the first two given equations gives us: $\begin{aligned} & \left( {{x}_{1}}+2{{x}_{2}}+{{x}_{3}} \right)+\left( 2{{x}_{1}}+3{{x}_{2}}+{{x}_{3}} \right)=3+3 \\\ & \Rightarrow 3{{x}_{1}}+5{{x}_{2}}+2{{x}_{3}}=6 \\\ \end{aligned}$ Here, we can see clearly that the above found expression is same as equation third of the given question in the left hand side as $3{{x}_{1}}+5{{x}_{2}}+2{{x}_{3}}=1$ but the right hand side of the both the equations are not same. This contradicts the system of linear equations rules that the same equation can’t yield different results. If the similar equations give different results then, the given system of linear equations has no solution. So, it is evident that the system has no solution.