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

Question

For a $3 	imes 3$ matrix $M$, let trace $(M)$ denote the sum of all the diagonal elements of $M$. Let $A$ be a $3 	imes 3$ 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 :

Accepted Answer

$$ B = \operatorname{adj}(\operatorname{adj}(2A)) = \det(2A) \cdot (2A) $$

Since $ A $ is a $ 3 \times 3 $ matrix with

$$ \det(A) = \frac{1}{2}, $$

the determinant of $ 2A $ is computed as

$$ \det(2A) = 2^3 \det(A) = 8 \cdot \frac{1}{2} = 4. $$

Thus,

$$ B = 4 \cdot (2A) = 8A. $$

Now, compute the determinant and the trace of $ B $:

Determinant of $ B $:

$$ \det(B) = \det(8A) = 8^3 \det(A) = 512 \cdot \frac{1}{2} = 256. $$

Trace of $ B $:

$$ \operatorname{trace}(B) = \operatorname{trace}(8A) = 8 \cdot \operatorname{trace}(A) = 8 \cdot 3 = 24. $$

Finally, adding these results:

$$ \det(B) + \operatorname{trace}(B) = 256 + 24 = 280. $$

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