For the system of linear equations α x+y+z=1,x+α y+z=1,x+y+α z=β, which one of the following statements is NOT correct?

Solution

Maths sir · 3 steps

Add all three equations: the sum x+y+z falls out, and the determinant decides each case.

1. Find the determinant. The coefficient determinant is |α, 1, 1; 1, α, 1; 1, 1, α| = α³ - 3α + 2. It factorises as below, so D = 0 only when α = 1 or α = -2.

   D = (α - 1)²(α + 2), (α + 2)(x + y + z) = 2 + β

2. Add the equations. Adding all three gives the second equation above. When D = 0, this sum and the equations themselves tell you whether the system is consistent.

   Option	What happens	Statement is
   (a) α = 1, β = 1	All three equations become x + y + z = 1, so there is one plane only	correct
   (b) α = 2, β = -1	D = 1² × 4 = 4 ≠ 0, so the solution is unique	not correct
   (c) α = 2, β = 1	4(x + y + z) = 3, so x + y + z = 3/4	correct
   (d) α = -2, β = 1	The sum gives 0 = 3, which is impossible	correct

3. So. At α = 2 the determinant is non-zero, so the system has exactly one solution for every β. Option (b) is the statement that is not correct.

The idea to keep. If D ≠ 0, the solution is unique, whatever the right-hand side is. Only when D = 0 do you check consistency, and adding the equations is often the fastest check.

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