CAIE
Further Paper 2
2023
November
Q1
4 marks
Standard +0.8
Show that the system of equations
$$14x - 4y + 6z = 5,$$
$$x + y + kz = 3,$$
$$-21x + 6y - 9z = 14,$$
where \(k\) is a constant, does not have a unique solution and interpret this situation geometrically. [4]