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MathematicsInteger+4Question 10 of 51

JEE Advanced 2024 · Paper 1, Mathematics question 10

Let S={A=(01c1ad1be):a,b,c,d,e∈{0,1} and ∣A∣∈{−1,1}}S = \left\{A = \begin{pmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{pmatrix} : a, b, c, d, e \in \{0,1\} \text{ and } |A| \in \{-1, 1\}\right\}, where ∣A∣|A| denotes the determinant of AA. Then the number of elements in SS is _____.

Answer and worked solution

Official answer: 16

Determinant (expand along the first row):
∣A∣=−1⋅(e−d)+c (b−a)=(d−e)+c (b−a).|A| = -1\cdot(e - d) + c\,(b - a) = (d - e) + c\,(b - a).

Case c=0c = 0. Need d−e=±1d - e = \pm1, i.e. d≠ed \ne e: 2 ways. a,ba, b are free: 4 ways. Total 88.

Case c=1c = 1. Need (d−e)+(b−a)=±1(d-e) + (b-a) = \pm1. Each difference is 00 (2 ways) or ±1\pm1 (1 way each).

  • Sum =1= 1: (1,0)(1,0) or (0,1)(0,1) gives 1⋅2+2⋅1=41\cdot2 + 2\cdot1 = 4 ways.
  • Sum =−1= -1: likewise 4 ways.

Total 88.

Answer: 8+8=168 + 8 = 16.

Topics:Determinants and propertiesFundamental principle of counting
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