Matrices & Determinants Question - None

Question

The set of all values of λ for which the system of linear equations
2x₁ - 2x₂ + x₃ = λx₁,
2x₁ - 3x2 + 2x₃ = λx₂,
-x₁ + 2x₂ = λx₃
has a non-trivial solution

Accepted Answer

The system can be written as:
(2-λ)x₁ - 2x₂ + x₃ = 0
2x₁ - (3+λ)x₂ + 2x₃ = 0
-x₁ + 2x₂ - λx₃ = 0
For a non-trivial solution, the determinant of the coefficient matrix must be zero:
| 2-λ -2 1 |
| 2 -(3+λ) 2 | = 0
| -1 2 -λ |
Expanding the determinant:
(2-λ)[(3+λ)(λ) - 4] + 2[-2λ + 2] + 1[4 - (3+λ)] = 0
(2-λ)(λ² + 3λ - 4) + 2(2 - 2λ) + (1 - λ) = 0
(2-λ)(λ+4)(λ-1) + 4(1-λ) + (1-λ) = 0
(1-λ) [-(2-λ)(λ+4) + 5] = 0
(1-λ) [-(λ² + 2λ - 8) + 5] = 0
(1-λ) [3 - 2λ - λ²] = 0
(1-λ) (-(λ² + 2λ - 3)) = 0
-(1-λ)(λ+3)(λ-1) = 0
(λ-1)²(λ+3) = 0
The values of λ are 1 and -3. Thus, the set contains two elements.

Question:

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