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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Matrices

NDA 2024 · Mathematics · Matrices

Let A and B be matrices of order 3 × 3 . If |A|= 1 2 2 and...

Let A and B be matrices of order 3×33 \times 3. If ∣A∣=122|A|=\frac{1}{2 \sqrt{2}} and ∣B∣=1729|B|=\frac{1}{729}, then what is the value of ∣2B(adj⁡(3A)∣\mid 2 \mathrm{ B}(\operatorname{adj}(3 \mathrm{ A}) \mid ?

Options

  1. A27
  2. B2722\frac{27}{2 \sqrt{2}}
  3. C272\frac{27}{2}
  4. D1

Correct answer

D. 1

Step-by-step solution

∣2B[adj⁡(3A)]∣=∣2B∣adj⁡(3A)∣=23∣B∣∣3A∣3−1=8∣B∣(33∣A∣)2=8∣B∣×36∣A∣2=8×1729×36×(122)2=8×1729×729×18=1\begin{aligned}|2 \mathrm{ B}[\operatorname{adj}(3 \mathrm{ A})]| & =|2 \mathrm{ B}| \operatorname{adj}(3 \mathrm{ A}) \mid \\ = & 2^3|\mathrm{ B}||3 \mathrm{ A}|^{3-1} \\ = & 8|\mathrm{ B}|\left(3^3|\mathrm{ A}|\right)^2 \\ = & 8|\mathrm{ B}| \times 3^6|\mathrm{ A}|^2 \\ = & 8 \times \frac{1}{729} \times 3^6 \times\left(\frac{1}{2 \sqrt{2}}\right)^2 \\ = & 8 \times \frac{1}{729} \times 729 \times \frac{1}{8}=1\end{aligned}
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