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Mathematics · Permutations and Combinations

JEE Advanced 2024 — Paper 1 — Question 10

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

Answer: 16

Numerical answer — enter this value.

Step-by-step solution

∣A∣=(e−d)+c(b−a)|A|=(e-d)+c(b-a) Now for ∣A∣=1|\mathrm{A}|=1 or -1 Case: 1 e−d=1\mathrm{e}-\mathrm{d}=1 or −1,c(b−a)=0-1 , \mathrm{c}(\mathrm{b}-\mathrm{a})=0

(c=0,b−a=1 or −1→(2)(c=1,b−a=0→(2)(c=0,b−a=0→(2)\begin{aligned} & (c=0, b-a=1 \text { or }-1 \rightarrow(2) \\& (c=1, b-a=0 \rightarrow(2) \\& (c=0, b-a=0 \rightarrow(2) \end{aligned}

2×(2+2+2)=122 \times(2+2+2)=12 Case: 2 e−d=0,c(b−a)=1e-d=0 , c(b-a)=1 or -1 ⇒e=d∣c=1, b−a=1\Rightarrow \mathrm{e}=\mathrm{d} | \mathrm{c}=1, \mathrm{~b}-\mathrm{a}=1 or -1

2×2=42 \times 2=4 Total possibility =12+4=16=12+4=16

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Mathematics
Chapter
Permutations and Combinations
Topic
Combinations

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