UPSC NDA I 2024 Mathematics Paper

To determine the value of $|2 B(\operatorname{adj}(3 A))|$, we need to utilize some properties of determinants and adjugate matrices.First, recall that for a square matrix $A$ of order $n$, the adjugate (or adjoint) of $A$, denoted as $\operatorname{adj}(A)$ , has a determinant related to the determinant of $A$. Specifically, for a $3 \times 3$ matrix, this relationship is:$\operatorname{adj}(A)=\det(A) \cdot A^{-1}$Thus, the determinant of the adjugate of $A$, for a $3 \times 3$ matrix, is:$\det(\operatorname{adj}(A))=\det(A)^{n-1}=\det(A)^2$Given $\det(A)=\;\frac{1}{2\sqrt{2}}$, we can find:$\det(\operatorname{adj}(3 A))=(\det(3 A))^2$Next, recall the determinant property for a scalar multiple of a matrix:$\det(k A)=k^n \cdot \det(A)$For $3 A$, we have:$\det(3 A)=3^3 \cdot \det(A)=27 \cdot \;\frac{1}{2\sqrt{2}}=\;\frac{27}{2\sqrt{2}}$ Thus:$\det(\operatorname{adj}(3 A))=\left(\;\frac{27}{2\sqrt{2}}\right)^2=\;\frac{729}{8}$Now, for the matrix $2 B$, use the scalar multiple property again:$\det(2 B)=2^3 \cdot \det(B)=8 \cdot \;\frac{1}{729}=\;\frac{8}{729}$We now combine these results to find:$\left|2 B(\operatorname{adj}(3 A))\right|=\det(2 B \cdot \operatorname{adj}(3 A))=\det(2 B) \cdot \det(\operatorname{adj}(3 A))$Therefore:$\left|2 B(\operatorname{adj}(3 A))\right|=\;\frac{8}{729} \cdot \;\frac{729}{8}=1$