JEE Advanced Mathematics Question (2023) — Solution

Step-by-step solution

Given, System of equations, x + 2 y + z = 7 x + α z = 11 2 x - 3 y + β z = γ Now finding  Δ = 1 2 1 1 0 α 2 - 3 β = 0 ⇒ 3 α - 2 β - 2 α - 3 = 0 ⇒ 7 α - 2 β = 3 ⇒ β = 1 2 7 α - 3 Now finding, Δ 3 = 1 2 7 1 0 11 2 - 3 γ Now equating,  Δ 3 = 0  we get, ⇒ 33 - 2 γ - 22 + 7 - 3 = 0 ⇒ γ = 28 And  Δ 1 = 7 2 1 11 0 α γ - 3 β   ⇒ Δ 1 = 21 α - 2 11 β - α γ - 33 ⇒ Δ 1 = 21 α - 22 β + 2 α γ - 33 Similarly,    Δ 2 = 1 7 1 1 11 α 2 γ β ⇒ Δ 2 = 11 β - α γ - 7 β - 2 α + γ - 22 ⇒ Δ 2 = 14 α + 4 β + γ - α γ - 22 P  If  β = 1 2 7 α - 3 and γ = 28 Δ = 0 ,   Δ 1 = 0 ,   Δ 2 = 0 ,   Δ 3 = 0 Infinitely many solutions x = 11 ,   y = - 2 and z = 0  will satisfy all the three given equations, so it is a solution. Q  If β = 1 2 7 α - 3 and γ ≠ 28  then Δ = 0 , but Δ 3 ≠ 0  so no solution R  If β ≠ 1 2 7 α - 3 ,   α = 1  and γ ≠ 28 Δ ≠ 0 ,   Δ 3 ≠ 0  so a unique solution S  If  β ≠ 1 2 7 α - 3 ,   α = 1 ,   γ = 28 Δ ≠ 0 ,   Δ 3 = 0 ,   Δ 1 ≠ 0 ,   Δ 2 ≠ 0 , so a unique solution x = 11 ,   y = - 2  and z = 0  will satisfy all the three equations Option A is correct.