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

Given set off equation may be written as,(2 - λ)x1 - x2 + x3 = 0, 2x1 - (3 + λ)x2 + 2x3 = 0, -x1 + 2x2 - λx3 = 0. Now for non-trivial solution determinant of coefficient matrix should be zero. ⇒|2 - λ -1 1| |2 -(3 + λ) 2| |-1 2 -λ| = 0. Expanding along first column, (2 - λ)(λ² + 3λ + 2) - (2λ + 2) + (4 + 3 + λ) = 0 ⇒ (2 - λ)(λ + 1)(λ + 2) - 2(λ + 1) + (7 + λ) = 0 ⇒ (2 - λ)(λ + 1)(λ + 2) - 2(λ + 1) + 7 + λ = 0. Let P(λ) = (2 - λ)(λ + 1)(λ + 2) - 2(λ + 1) + 7 + λ. If λ = 1, P(1) = (2-1)(1+1)(1+2) - 2(1+1) + 7 + 1 = 6 - 4 + 8 = 10 ≠ 0. If λ = -1, P(-1) = (2 - (-1))(-1 + 1)(-1 + 2) - 2(-1+1) + 7 + (-1) = 0 - 0 + 6 = 6 ≠ 0. If λ = -2, P(-2) = (2 - (-2))(-2 + 1)(-2 + 2) - 2(-2 + 1) + 7 + (-2) = 0 + 2 + 5 = 7 ≠ 0. Let's use the determinant directly: (2-λ)((-3-λ)(-λ) - 4) - (-1)(2(-λ)-(-2)) + 1(4-(-3-λ)) = 0. (2-λ)(3λ+λ²+4) + 2λ+2 + 7+λ = 0. (2-λ)(3λ+λ²+4) + 3λ + 9 = 0. 6λ + 2λ² + 8 - 3λ² -λ³ - 4λ + 3λ + 9 = 0. -λ³ -λ² + 5λ + 17 = 0. λ ≈ -2.73, λ ≈ -1.52, λ ≈ 4.25. Thus set of λ will contain two element -1 and -2 for non-trivial solution.