JEE Advanced 2024 · Paper 1 · Mathematics Q14

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MathematicsList match+3 / -1Question 14 of 51

Let $\alpha$ and $\beta$ be the distinct roots of the equation $x^2 + x - 1 = 0$ . Consider the set $T = \{1, \alpha, \beta\}$ . For a $3\times 3$ matrix $M = (a_{ij})_{3\times 3}$ , define $R_i = a_{i1} + a_{i2} + a_{i3}$ and $C_j = a_{1j} + a_{2j} + a_{3j}$ for $i = 1, 2, 3$ and $j = 1, 2, 3$ .

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

The correct option is

List-I

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
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
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 $\left\{\begin{pmatrix} x \\ y \\ z \end{pmatrix} : x, y, z \in \mathbb{R},\ M\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_{12} \\ 0 \\ -a_{23} \end{pmatrix}\right\}$ is
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

List-II

1
1
2
12
3
infinite
4
6
5
0

P → 4; Q → 2; R → 5; S → 1

P → 2; Q → 4; R → 1; S → 5

P → 2; Q → 4; R → 3; S → 5

P → 1; Q → 5; R → 3; S → 4

Answer and worked solution

Official answer: C

$\alpha + \beta = -1$ , so $1 + \alpha + \beta = 0$ . Three entries from $T$ add up to zero only when they are $1, \alpha, \beta$ in some order (check: $3t = 0$ and $2t + s = 0$ have no solutions in $T$ ).

(P) Every row and every column is a permutation of $(1,\alpha,\beta)$ , so $M$ is a $3\times3$ Latin square: $12$ . (P)→(2).

(Q) Symmetric with zero column sums, so rows are permutations too: a symmetric Latin square of order 3. Its diagonal holds each symbol once (6 orders), and that fixes the rest: $6$ . (Q)→(4).

(R) $M = \begin{pmatrix}0 & -a_{21} & -a_{31} \\ a_{21} & 0 & -a_{32} \\ a_{31} & a_{32} & 0\end{pmatrix}$ and the right side is $(-a_{21}, 0, a_{32})^T = M\,(0,1,0)^T$ . The system is consistent, and a $3\times3$ skew-symmetric matrix is singular, so there are infinitely many solutions. (R)→(3).

(S) $M(1,1,1)^T = 0$ , so $|M| = 0$ . (S)→(5).

Option (C).

Topics:Special matrices Systems of linear equations Roots and coefficients

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JEE Advanced 2024 · Paper 1, Mathematics question 14