Find the set of values of a for which the system of equations x + 4y + 12z = 5, 2x + ay + 12z = a − 1, 3x + 12y + 2az = 10, has a unique solution.

Mark scheme

Marking points from the official mark scheme (May/June 2010). The mark or mark type is shown in brackets.

1. [2] (a) Calculate the determinant of the coefficient matrix A: det(A) = 1(a(2a)-12(12)) - 4(2(2a)-12(3)) + 12(2(12)-a(3)) = 2a²-144 - 4(4a-36) + 12(24-3a) = 2a²-144 - 16a+144 + 288-36a = 2a²-52a+288 = 2(a²-26a+144) = 2(a-8)(a-18).
2. [2] (a) For a unique solution, det(A) ≠ 0. Therefore, 2(a-8)(a-18) ≠ 0, which means a ≠ 8 and a ≠ 18. (A1 for each condition, or B1 for 2(a-8)(a-18)=0, then A1 for conditions)
3. [2] (b) When a = 18, the augmented matrix (or equivalent row operations) leads to an inconsistency. For example, using row operations: [[1, 4, 12 | 5], [2, 18, 12 | 17], [3, 12, 36 | 10]]. R2-2R1: [0, 10, -12 | 7]. R3-3R1: [0, 0, 0 | -5]. The last row 0x+0y+0z=-5 is impossible, showing no solution.
4. [1] (c) When a = 8, the system becomes: x+4y+12z = 5, 2x+8y+12z = 7, 3x+12y+16z = 10. Perform row operations on the augmented matrix: [[1, 4, 12 | 5], [2, 8, 12 | 7], [3, 12, 16 | 10]]. R2-2R1: [0, 0, -12 | -3]. R3-3R1: [0, 0, -20 | -5]. From R2': -12z = -3 => z = 1/4. From R3': -20z = -5 => z = 1/4 (consistent). This means the system reduces to two independent equations, for example, x + 4y = 2 and z = 1/4.
5. [1] (c) Since there are more variables than independent equations, there are infinitely many solutions. Parameterise solutions: Let y = λ. Then x = 2 - 4λ. So the solutions are of the form x = 2 - 4λ, y = λ, z = 1/4 (where λ is a real number).
6. [1] (c) Since a parameter can take any real value, there are an infinite number of solutions.
7. [2] (c) Substitute into x + y + z = 1: (2 - 4λ) + λ + 1/4 = 1. => 2 - 3λ + 1/4 = 1. => 9/4 - 3λ = 1. => 3λ = 5/4. => λ = 5/12. Thus, x = 2 - 4(5/12) = 2 - 5/3 = 1/3. y = 5/12. z = 1/4. The particular solution is (1/3, 5/12, 1/4).