QJEEM25Ma5ad1: JEE Main 2025 Mathematics - Class 12 JEEM25Ma5ad1 Solution: For a matrix , let trace denote the sum of all the diagonal…

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

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 $B$ :

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

Finally, adding these results:

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