Question

The system of algebraic equations given below has

x  + 2y + z = 4

2x + y + 2z = 5

x – y + z = 1

1. A unique solution of x = 1, y = 1 and z = 1
2. Only the two solutions of (x = 1, y = 1 and z = 1) and (x = 2, y = 1 and z = 0)
3. Infinite number of solutions
4. No feasible solution

Answer 1

Correct Answer - Option 3 : Infinite number of solutions

Concept:

Let

[A] is the Coefficient matrix

[A/B] be Augmented matrix

n = total number of variables

Case 1: ρ(A) = ρ(A/B) = n

In this case, the system will be consistent and will have a unique solution.

Case 2: ρ(A) = ρ(A/B) < n

In this case, the system will be consistent and will have infinite solutions.

Case 3: ρ(A) < ρ(A/B)

In this case, the system will be inconsistent and will have no solution.

Calculation:

Given:

x  + 2y + z = 4

2x + y + 2z = 5

x – y + z = 1

Here n = 3

Augmented matrix is:

\(\left[ {A/B} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 1&2&1\\ 2&1&2\\ 1&{ - 1}&1 \end{array}} \right|\begin{array}{*{20}{c}} 4\\ 5\\ 1 \end{array}} \right]\)

R₂ → R₂ – 2R₁ and R₃ → R₃ – R₁

_{\(\left[ {A/B} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 1&2&1\\ 0&{ - 3}&0\\ 0&{ - 3}&0 \end{array}} \right|\begin{array}{*{20}{c}} {\;\;\;4}\\ { - 3}\\ { - 3} \end{array}} \right]\)}

R₃ → R₃ – R₂

\($\left[ {A/B} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 1&2&1\\ 0&{ - 3}&0\\ 0&0&0 \end{array}} \right|\begin{array}{*{20}{c}} {\;\;4}\\ { - 3}\\ {\;\;0} \end{array}} \right]\)

ρ(A) = ρ(A/B)  = 2 < 3

∴ system will be consistent and will have infinite solutions.