Q26 / 75 · JEE Main 2025 (22 Jan Shift 2) · Mathematics

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

  • A
    56
  • B
    132
  • C
    174
  • Correct answer
    D
    280

Solution

$\because \operatorname{tr}(A)=3 \text { and }|A|=\frac{1}{2}$ Now, $B=\operatorname{adj}(\operatorname{adj}(2 A))=|2 A|^{3-2} \cdot(2 A)$ $=2^3|A| \cdot 2 A=8 A$ $\begin{aligned} & \therefore \operatorname{tr}(B)=8 \operatorname{tr}(A)=24 \\ & \text { and }|B|=|8 A|=8^3 \cdot \frac{1}{2}=256 \\ & \therefore \quad \operatorname{trace}(B)+|B|=24+256\end{aligned}$ $=280$

Hence, the correct answer is option: 'D'.

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