JEE Advanced 2023 — Paper 1 — MATH-IV solution

JEE Advanced2023Paper 1MATH-IV

Let $α$ , $β$ and $γ$ be real numbers. Consider the following system of linear equations

$x + 2 y + z = 7$

$x + α z = 11$

$2 x - 3 y + β z = γ$

Match each entry in List-I to the correct entries in List-II.

List-I	List-II
(P) If $β = \frac{1}{2} (7 α - 3)$ and $γ = 28$ , then the system has	(1) a unique solution
(Q) If $β = \frac{1}{2} (7 α - 3)$ and $γ \neq = 28$ , then the system has	(2) no solution
(R) If $β \neq = \frac{1}{2} (7 α - 3)$ where $α = 1$ and $γ \neq = 28$ , then the system has	(3) infinitely many solutions
(S) If $β \neq = \frac{1}{2} (7 α - 3)$ where $α = 1$ and $γ = 28$ , then the system has	(4) $x = 11$ , $y = - 2$ and $z = 0$ as a solution
	(5) $x = - 15$ , $y = 4$ and $z = 0$ as a solution

A
(P) $\to$ (3); (Q) $\to$ (2); (R) $\to$ (1); (S) $\to$ (4)
B
(P) $\to$ (3); (Q) $\to$ (2); (R) $\to$ (5); (S) $\to$ (4)
C
(P) $\to$ (2); (Q) $\to$ (1); (R) $\to$ (4); (S) $\to$ (5)
D
(P) $\to$ (2); (Q) $\to$ (1); (R) $\to$ (1); (S) $\to$ (3)

Solution

The coefficient determinant is

$Δ = 121 10 α 2 - 3 β = 7 α - 2 β - 3$ .

So $Δ = 0 ⟺ β = \frac{1}{2} (7 α - 3)$ .

Similarly compute $Δ_{x}$ , $Δ_{y}$ , $Δ_{z}$ (by replacing each column with the RHS column $(7, 11, γ)^{T}$ ). Each of these three determinants simplifies and becomes zero precisely when $γ = 28$ .

(P) $Δ = 0$ and all $Δ_{x}, Δ_{y}, Δ_{z} = 0$ : by Cramer's analysis the system is consistent with $Δ = 0$ , hence infinitely many solutions (3).

(Q) $Δ = 0$ but at least one $Δ_{i} \neq = 0$ : the system is inconsistent, hence no solution (2).

(R) $Δ \neq = 0$ (with $α = 1$ , $γ \neq = 28$ ): the system has a unique solution (1).

(S) $Δ \neq = 0$ with $α = 1$ , $γ = 28$ : the unique solution can be checked by substitution. Trying $(x, y, z) = (11, - 2, 0)$ in the original equations: $11 - 4 + 0 = 7$ $✓$ ; $11 + 0 = 11$ $✓$ ; $22 + 6 + 0 = 28$ $✓$ . So (4).

Mapping: P $\to$ 3, Q $\to$ 2, R $\to$ 1, S $\to$ 4. The correct option is (A).

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