Mathematics · Permutations and Combinations

JEE Advanced 2023 — Paper 2 — Question 12

Let R={(a3bc2d050):a,b,c,d∈{0,3,5,7,11,13,17,19}}R=\left\{\left(\begin{array}{lll}a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0\end{array}\right): a, b, c, d \in\{0,3,5,7,11,13,17,19\}\right\}. Then the number of invertible matrices in R is

Answer: 3780

Numerical answer — enter this value.

Step-by-step solution

Total matrices =84=4096=8^{4}=4096 ∣R∣=5(bc−ad)|\mathrm{R}|=5(\mathrm{bc}-\mathrm{ad}) No. of non-invertible matrices:

bc=ad\mathrm{bc}=\mathrm{ad}

Case I: if a, b, c, d≠0d \neq 0, then cases =7C2⋅2!2!+7C1(1)=91={ }^{7} C_{2} \cdot 2!2!+{ }^{7} C_{1}(1)=91 Case II: if ad = bc =0=0, then cases =15C1⋅15C1=225={ }^{15} \mathrm{C}_{1} \cdot{ }^{15} \mathrm{C}_{1}=225 (ad &\& bc can take any combination from 0×0,0×3,0×5,…,0×19,3×0,5×0,…19×00 \times 0,0 \times 3,0 \times 5, \ldots, 0 \times 19,3 \times 0,5 \times 0, \ldots 19 \times 0 ) No. of invertible matrices =4096−(91+225)=3780=4096-(91+225)=3780

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 2
Subject
Mathematics
Chapter
Permutations and Combinations
Topic
Combinations