"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 :"

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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:

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

"

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 :

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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 :

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