Question

Consider the following system of equations

2x₁ + x₂ + x₃ = 0,

x₂ – x₃ = 0,

x₁ + x₂ = 0,

This system has

1. A unique solution
2. No solution
3. Infinite number of solutions
4. Five solutions

Answer 1

Correct Answer - Option 3 : Infinite number of solutions

Concept:

For a Homogenous system, AX = O

[A] is the Coefficient matrix

[A/O] be Augmented matrix

[O] is a null matrix and

n = total number of variables

Case 1: ρ(A) = ρ(A/O) = n

In this case, the system possesses only a zero solution (or Trivial solution) i.e unique solution.

Case 2: ρ(A) = ρ(A/O) < n

In this case, the system has an infinite number of non-zero solutions (or Trivial solutions).

Case 3: ρ(A) = ρ(A/O)

Hence, inconsistency does not arise, moreover, zero solution is always a solution to it.

Calculation:

Given:

2x1 + x2 + x3 = 0

x2 – x3 = 0

x1 + x2 = 0

Here n = 3

Now, we know that

For a Homogenous system, AX = O

Augmented matrix is:

\(\left[ {A/O} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 2&1&1\\ 0&1&-1\\ 1&{ 1}&0 \end{array}} \right|\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right]\)

R₃ → R₃ - (R₁/2)

\(\left[ {\left. {\begin{array}{*{20}{c}} 2&1&1\\ 0&1&{ - 1}\\ 0&{1/2}&{-1/2} \end{array}} \right|\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right]\)

R3 → R3 - (R₂/2)

\(\left[ {\left. {\begin{array}{*{20}{c}} 2&1&1\\ 0&1&{ - 1}\\ 0&0&0 \end{array}} \right|\begin{array}{*{20}{c}} 0\\ 0\\ 0 \end{array}} \right]\)

As, ρ(A) = ρ(A/O) = 2 < 3

∴ The system is consistent and will have infinite number of solutions.