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JEE Advanced 2024 Paper 1

Section: Mathematics
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Question : 14 of 51
Marks: +1, -0
Let α\alpha and β\beta be the distinct roots of the equation x2+x1=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)3×3)M= (a_{i j}) 3 \times 3), define Ri=ai1+ai2+aβR_i=a_{i 1}+a_{i 2}+a_\beta and Cj=a1j+a2j+a3jC_j=a_{1 j}+a_{2 j}+a_{3 j} 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    List-II
 (P)  The number of matrices M=(aij)3×3M= (a_{i j}) 3 \times 3 with all entries in TT such that Ri=Cj=0R_i=C_j=0 for all i,ji, j is  (1)  1
 (Q)  The number of symmetric matrices M=(aij)3×3M= (a_{i j}) 3 \times 3 with all entries in TT such that Cj=0C_j=0 for all jj is  (2)  12
 (R)  Let M=(aij)3×3M= (a_{i j}) _{3 \times 3} be a skew symmetric matrix such that aijTa_{ij} \in T for i>ji>j . Then the number of elements in the set
{(xyz):x,yzR,M(xyz)=(a120a23)}\{ \begin{pmatrix} x \\ y \\ z \end{pmatrix} : x, y \cdot z \in R, M\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_{12} \\ 0 \\ -a_{23} \end{pmatrix} \} is
 (3)  Infinite
 (S)  Let M=(aij)3×3M= (a_{i j}) _{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  (4)  6
     (5)  0
The correct option is
[JEE Adv 2024 P1]
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