Linear system solvable for which λ

The system is x + y + z = 1, x + 4y + 10z = λ², x + 2y + 4z = λ. It has a solution only if rank A = rank [A|B].

[A|B] = [1 1  1 |  1 ]
        [1 4 10 | λ² ]
        [1 2  4 |  λ ]

Step 1. R₂ → R₂ − R₁, R₃ → R₃ − R₁:

[1 1 1 |   1  ]
[0 3 9 | λ²−1 ]
[0 1 3 | λ−1  ]

Step 2. R₂ → R₂ − 3R₃:

[1 1 1 |          1          ]
[0 0 0 | λ²−1−3λ+3 = λ²−3λ+2 ]
[0 1 3 |         λ−1         ]

Interchange R₂ and R₃:

[1 1 1 |      1       ]
[0 1 3 |     λ−1      ]
[0 0 0 | (λ−1)(λ−2)   ]

Condition

rank A = 2 always. The system is consistent only if rank [A|B] = 2, i.e. (λ − 1)(λ − 2) = 0.

The equations have a solution only for λ = 1 or λ = 2. In both cases rank = 2 < 3, so there are infinitely many solutions (one free variable z = k).

Case λ = 1

• y + 3z = 0 ⇒ y = −3k
• x = 1 − y − z = 1 + 3k − k = 1 + 2k

x = 1 + 2k, y = −3k, z = k.

Case λ = 2

• y + 3z = 1 ⇒ y = 1 − 3k
• x = 1 − y − z = 1 − 1 + 3k − k = 2k

x = 2k, y = 1 − 3k, z = k.

For any other value of λ, the system is inconsistent (no solution).