JEE Advanced 2024 — Paper 1 — MATH-IV solution

JEE Advanced2024Paper 1MATH-IV

Let $α$ and $β$ be the distinct roots of the equation $x^{2} + x - 1 = 0$ . Consider the set $T = {1, α, β}$ . For a $3 \times 3$ matrix $M = (a_{ij})_{3 \times 3}$ , define $R_{i} = a_{i 1} + a_{i 2} + a_{i 3}$ and $C_{j} = a_{1 j} + a_{2 j} + a_{3 j}$ for $i = 1, 2, 3$ and $j = 1, 2, 3$ .

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

List-I	List-II
(P) The number of matrices $M = (a_{ij})_{3 \times 3}$ with all entries in $T$ such that $R_{i} = C_{j} = 0$ for all $i, j$ is	(1) 1
(Q) The number of symmetric matrices $M = (a_{ij})_{3 \times 3}$ with all entries in $T$ such that $C_{j} = 0$ for all $j$ is	(2) 12
(R) Let $M = (a_{ij})_{3 \times 3}$ be a skew symmetric matrix such that $a_{ij} \in T$ for $i > j$ . Then the number of elements in the set $⎩ ⎨ ⎧ x y z : x, y, z \in R, M x y z = a_{12} 0 - a_{23} ⎭ ⎬ ⎫$ is	(3) Infinite
(S) Let $M = (a_{ij})_{3 \times 3}$ be a matrix with all entries in $T$ such that $R_{i} = 0$ for all $i$ . Then the absolute value of the determinant of $M$ is	(4) 6
	(5) 0

The correct option is

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

Solution

From $x^{2} + x - 1 = 0$ : $α + β = - 1$ and $α β = - 1$ . Key observation: the only way three elements from $T = {1, α, β}$ sum to $0$ is to use one of each (since $1 + α + β = 1 + (- 1) = 0$ ).

(P) Each row must be a permutation of ${1, α, β}$ . For column sums to also be $0$ , each column must likewise be a permutation. This is a $3 \times 3$ Latin square count over three symbols: $3! \times 2 \times 1 = 12$ . So (P) $\to$ (2).

(Q) For a symmetric matrix with each column summing to $0$ , by symmetry each row also sums to $0$ . The diagonal triple must be a permutation of ${1, α, β}$ (giving $3! = 6$ choices), and once the diagonal is fixed, the off-diagonal entries are forced by row-sum and symmetry. Count: $6$ . So (Q) $\to$ (4).

(R) A real $3 \times 3$ skew-symmetric matrix has determinant $0$ . The augmented system is consistent (the given RHS lies in the column space by construction of skew-symmetric structure), so the system has infinitely many solutions. So (R) $\to$ (3).

(S) Each row of $M$ is a permutation of ${1, α, β}$ . Performing $C_{1} \to C_{1} + C_{2} + C_{3}$ makes the first column entirely zero (sum is $0$ in each row), so $∣ M ∣ = 0$ . So (S) $\to$ (5).

$∴ Option (C)$

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