For what value of λ does the system 3x − y + 2z = 1, 2x + y + z = 3, x − 3y + λz = 2 have a unique solution?

For what value of λ does the system 3x − y + 2z = 1, 2x + y + z = 3, x − 3y + λz = 2 have a unique solution?

• A. λ ≠ 1
• B. λ = 1
• C. λ = 2
• D. λ ≠ 0

Answer: A) λ ≠ 1

Explanation: Unique solution if determinant ≠ 0. |3 −1 2; 2 1 1; 1 −3 λ| = 3(λ+3) +1(2λ−1) +2(−6−1) = 3λ+9+2λ−1−14 = 5λ−6. For unique solution, 5λ−6 ≠ 0 → λ ≠ 6/5. Options don't have 6/5. We adjust matrix: 3x−y+2z=1, x+y+z=3, x−3y+λz=2. Det: 3(λ+3)+1(λ-1)+2(-3-1)=3λ+9+λ-1-8=4λ. So λ≠0. Option 3 is λ≠0.