Question

If the system of linear equations
\[\begin{align}
  & {{x}_{1}}+2{{x}_{2}}+3{{x}_{3}}=6 \\
 & {{x}_{1}}+3{{x}_{2}}+5{{x}_{3}}=9 \\
 & 2{{x}_{1}}+5{{x}_{2}}+a{{x}_{3}}=b \\
\end{align}\]
Is it consistent and having an infinite number of solutions, then?
 \[\begin{align}
  & \text{(A) a}\in \text{R- }\!\!\{\!\!\text{ 8 }\!\!\}\!\!\text{ and b}\in \text{R- }\!\!\{\!\!\text{ 15 }\!\!\}\!\!\text{ } \\
 & \text{(B) a=8, b can be any real number} \\
 & \text{(C) a=8, b=15} \\
 & \text{(D) b=15, a can be any real number} \\
\end{align}\]

Answer 1

Hint: We should know that a system of linear equations is said to be consistent and have an infinite number of solutions if the number of unknowns is greater than the number of linear equations. If the number of unknowns are less than the number of linear equations, then we can say that the system of linear equations is said to be inconsistent. By using this concept, we should solve the given problem. From the question, we were given a set of linear equations. Now this set has three linear equations but thus the set should represent only two linear equations. So, let us consider that by adding two equations of this set will result in another equation. Now by comparing the two equations, we can get the value of a and b. In other cases, if the three linear equations represent the same equation then also the system will be consistent and have an infinite number of solutions.

Complete step-by-step answer:
Before solving the question, we should know that a system of linear equations is said to be consistent and has an infinite number of solutions if the number of unknowns is greater than the number of linear equations.

Now we should apply this concept to solve this problem.
From the question, we were given that
\[\begin{align}
  & {{x}_{1}}+2{{x}_{2}}+3{{x}_{3}}=6 \\
 & {{x}_{1}}+3{{x}_{2}}+5{{x}_{3}}=9 \\
 & 2{{x}_{1}}+5{{x}_{2}}+a{{x}_{3}}=b \\
\end{align}\]
Now let us assume
\[\begin{align}
  & {{x}_{1}}+2{{x}_{2}}+3{{x}_{3}}=6.....(1) \\
 & {{x}_{1}}+3{{x}_{2}}+5{{x}_{3}}=9......(2) \\
 & 2{{x}_{1}}+5{{x}_{2}}+a{{x}_{3}}=b.....(3) \\
\end{align}\]
From equation (1), equation (2) and equation (3) we should find the total number of variables collectively.
From equation (1), equation (2) and equation (3), the variables are \[{{x}_{1}},{{x}_{2}},{{x}_{3}}\].
So, we can say that the number of variables in equation (1), equation (2) and equation (3) collectively is equal to 3.
So, to have consistent and infinite solutions the number of equations in the linear set should be less than the number of variables.
So, in the given linear set we have to have 2 linear equations or one linear equation.
Case 1: (the given set of equations have only two linear equations)
In this case, we should get an equation which is the sum of other two equations such that three linear equations belong to the same linear set of equations.
Now let us assume equation (3) as sum of equation (1) and equation (2).
Now let us add equation (1) and equation (2).
\[\begin{align}
  & \Rightarrow ({{x}_{1}}+2{{x}_{2}}+3{{x}_{3}})+({{x}_{1}}+3{{x}_{2}}+5{{x}_{3}})=9+6 \\
 & \Rightarrow 2{{x}_{1}}+5{{x}_{2}}+8{{x}_{3}}=15......(4) \\
\end{align}\]
Now we will compare equation (4) with equation (3), we get
\[a=8;b=15\]
 Case 2: (the given set of equations has only one linear equation)
By observing equation (1) and equation (2), we can simply say that they are minimum two equations in the given set. But we are having three variables, so we can say that this case is not possible at all.

So, the correct answer is “Option C”.

Note: The sum can be solved in an alternative method as well.
We know that for a set of three linear equations,
\[\begin{align}
  & {{a}_{1}}{{x}_{1}}+{{b}_{1}}{{x}_{2}}+{{c}_{1}}{{x}_{3}}={{d}_{1}} \\
 & {{a}_{2}}{{x}_{1}}+{{b}_{2}}{{x}_{2}}+{{c}_{2}}{{x}_{3}}={{d}_{2}} \\
 & {{a}_{3}}{{x}_{1}}+{{b}_{3}}{{x}_{2}}+{{c}_{3}}{{x}_{3}}={{d}_{3}} \\
\end{align}\]
The solutions are said to be consistent if the value of \[\left( \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right)=0\].
The given linear set of equations has infinite solutions if \[\left( \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{d}_{3}} \\
   {{a}_{2}} & {{b}_{2}} & {{d}_{3}} \\
   {{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\
\end{matrix} \right)=0\].
In the question, we were given that the system of linear equations
\[\begin{align}
  & {{x}_{1}}+2{{x}_{2}}+3{{x}_{3}}=6 \\
 & {{x}_{1}}+3{{x}_{2}}+5{{x}_{3}}=9 \\
 & 2{{x}_{1}}+5{{x}_{2}}+a{{x}_{3}}=b \\
\end{align}\]
Is consistent and has an infinite number of solutions.
Now we have to apply the conditions for the linear set to be consistent.
\[\begin{align}
  & \left( \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 2 & 3 \\
   1 & 3 & 5 \\
   2 & 5 & a \\
\end{matrix} \right)=0 \\
 & \Rightarrow 1(3a-25)-2(a-10)+3(5-6)=0 \\
 & \Rightarrow a=8....(1) \\
\end{align}\]
Now we have to apply the conditions for a linear set to have infinite solutions.
\[\begin{align}
  & \left( \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{d}_{3}} \\
   {{a}_{2}} & {{b}_{2}} & {{d}_{3}} \\
   {{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 2 & 6 \\
   1 & 3 & 9 \\
   2 & 5 & b \\
\end{matrix} \right)=0 \\
 & \Rightarrow 1(3b-45)-2(b-18)+6(5-6)=0 \\
 & \Rightarrow b=15....(2) \\
\end{align}\]
From equation (1) and equation (2), it is clear that option C is correct.