Find the value of a for which the system of equations x - y + 2z = 4, x + ay - 3z = b, x - y + 7z = 13, where a and b are constants, has no unique solution. Taking a as the value just found, (i) find the general solution in the case b = –5, (ii) interpret the situation geometrically in the case b ≠ -5.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2014). The mark or mark type is shown in brackets.

1. [M1] For no unique solution, the determinant of the coefficient matrix must be zero: det(A) = 0.
2. [A1A1] det(A) = 1(7a - 3) + 1(7 - (-3)) + 2(-1 - a) = 7a - 3 + 10 - 2 - 2a = 5a + 5. So, 5a + 5 = 0 ⇒ a = -1.
3. [M1] Set a = -1 and b = -5. Row operations on the augmented matrix: ((1, -1, 2, 4), (1, -1, -3, -5), (1, -1, 7, 13)) → ((1, -1, 2, 4), (0, 0, -5, -9), (0, 0, 5, 9)) → ((1, -1, 2, 4), (0, 0, 5, 9), (0, 0, 0, 0)).
4. [A1] This gives the equations x - y + 2z = 4 and 5z = 9.
5. [M1A1] From 5z = 9, z = 9/5. Substitute into the first equation: x - y + 2(9/5) = 4 ⇒ x - y + 18/5 = 4 ⇒ x - y = 2/5. Let y = t (parameter). Then x = t + 2/5. The general solution is (x, y, z) = (t + 2/5, t, 9/5) for any real t.
6. [B1] In the case b ≠ -5 (with a=-1), the system has no solution. Geometrically, the planes form a prism (or are parallel and distinct, resulting in no common intersection point).