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JEE Main 202522 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Matrices

For a 3 × 3 matrix M , let trace (M) denote the sum of all the...

For a 3×33 \times 3 matrix MM, let trace (M)(M) denote the sum of all the diagonal elements of MM. Let AA be a 3×33 \times 3 matrix such that ∣A∣=12|A|=\frac{1}{2} and trace (A)=3(A)=3. If B=adj⁡(adj⁡(2A))B=\operatorname{adj}(\operatorname{adj}(2 A)), then the value of ∣B∣+|B|+ trace (B) equals :

Options

  1. A56
  2. B132
  3. C174
  4. D280

Correct answer

D. 280

Step-by-step solution

\because \operatorname{tr}(A)=3 \text{ and }|A|=\frac{1}{2} Now, B=adj⁡(adj⁡(2A))=∣2A∣3−2⋅(2A)=23∣A∣⋅2A=8AB=\operatorname{adj}(\operatorname{adj}(2 A))=|2 A|^{3-2} \cdot(2 A) =2^3|A| \cdot 2 A=8 A∴tr⁡(B)=8tr⁡(A)=24 and ∣B∣=∣8A∣=83⋅12=256∴trace⁡(B)+∣B∣=24+256\begin{aligned} & \therefore \operatorname{tr}(B)=8 \operatorname{tr}(A)=24 \\ & \text{ and }|B|=|^{8} A|=8^3 \cdot \frac{1}{2}=256 \\ & \therefore \quad \operatorname{trace}(B)+|B|=24+256\end{aligned}=280=280
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