JEE Advanced 2024 · Paper 1

JEE Advanced 2024 Mathematics question on Matrices and Determinants

Let α\alpha and β\beta be the distinct roots of the equation x2+x−1=0x^2 + x - 1 = 0. Consider the set T={1,α,β}T = \{1, \alpha, \beta\}. For a 3×33 \times 3 matrix M=(aij)3×3M = (a_{ij})_{3\times 3}, define Ri=ai1+ai2+ai3R_i = a_{i1} + a_{i2} + a_{i3} and Cj=a1j+a2j+a3jC_j = a_{1j} + a_{2j} + a_{3j} for i=1,2,3i = 1,2,3 and j=1,2,3j = 1,2,3.

Match each entry in List-I to the correct entry in List-II.
List-I
(P) The number of matrices M=(aij)3×3M = (a_{ij})_{3\times 3} with all entries in TT such that Ri=Cj=0R_i = C_j = 0 for all i,ji,j, is
(Q) The number of symmetric matrices M=(aij)3×3M = (a_{ij})_{3\times 3} with all entries in TT such that Cj=0C_j = 0 for all jj, is
(R) Let M=(aij)3×3M = (a_{ij})_{3\times 3} be a skew symmetric matrix such that aij=0a_{ij} = 0 for i>ji > j. Then the number of elements in the set {(xyz):x,y,z∈R,M(xyz)=(a120−a23)}\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=(aij)3×3M = (a_{ij})_{3\times 3} be a matrix with all entries in TT such that Ri=0R_i = 0 for all ii. Then the absolute value of the determinant of MM is
List-II
(1) 1
(2) 12
(3) infinite
(4) 6
(5) 0
  1. A.
    (P)→(4)(Q)→(4)(R)→(5)(S)→(1)(P)\to(4)\quad (Q)\to(4)\quad (R)\to(5)\quad (S)\to(1)
  2. B.
    (P)→(2)(Q)→(4)(R)→(1)(S)→(5)(P)\to(2)\quad (Q)\to(4)\quad (R)\to(1)\quad (S)\to(5)
  3. C.
    (P)→(2)(Q)→(4)(R)→(3)(S)→(5)(P)\to(2)\quad (Q)\to(4)\quad (R)\to(3)\quad (S)\to(5)
  4. D.
    (P)→(1)(Q)→(5)(R)→(3)(S)→(4)(P)\to(1)\quad (Q)\to(5)\quad (R)\to(3)\quad (S)\to(4)
Show answer

Correct option: C

[JEE Advanced 2024 · Paper 1]

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