Let the system of linear equations
x + y + αz = 2
3x + y + z = 4
x + 2z = 1
have a unique solution (x *, y *, z *). If (α, x *), (y *, α) and (x *, -y *) are collinear points, then the sum of absolute values of all possible values of α is

Solution

Maths sir · 4 steps

Solve the system first: it gives (1, 1, 0) for every allowed α, so collinearity becomes one equation in α.

1. Unique solution condition. The coefficient determinant is |1, 1, α; 3, 1, 1; 1, 0, 2| = 2 - 5 - α = -(3 + α). A unique solution needs this to be non-zero, so α ≠ -3.
2. Solve the system. The third equation gives x = 1 - 2z. Put it in the second: y = 1 + 5z. The first then becomes (3 + α)z = 0, so z = 0 because α ≠ -3. Hence x^* = 1, y^* = 1, z^* = 0.
3. Collinearity condition. The points are (α, 1), (1, α) and (1, -1). Three points are collinear exactly when the triangle they form has zero area.

   |α, 1, 1; 1, α, 1; 1, -1, 1| = α(α + 1) - 0 + (-1 - α) = α² - 1 = 0

4. So. α = 1 or α = -1, and neither is -3, so both are allowed. The sum of absolute values is |1| + |-1| = 2, which is option (c).

The idea to keep. Find the solution before you touch the condition on it. And for collinearity, use the area determinant: it also catches the case where two of the points coincide.

Answer from the NTA final answer key. Solution by PrepWiser.