UPSC NDA I 2024 Mathematics Paper

To determine the value of |2B(adj(3A))|, 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 adj(A) , has a determinant related to the determinant of A. Specifically, for a 3×3 matrix, this relationship is:
adj(A)=det(A)⋅A−1
Thus, the determinant of the adjugate of A, for a 3×3 matrix, is:
det(adj(A))=det(A)n−1=det(A)2
Given det(A)=‌

1

2√2

, we can find:
det(adj(3A))=(det(3A))2
Next, recall the determinant property for a scalar multiple of a matrix:
det(kA)=kn⋅det(A)
For 3A, we have:
det(3A)=33⋅det(A)=27⋅‌

1

2√2

=‌

27

2√2

Thus:
det(adj(3A))=(‌

27

2√2

)2=‌

729

8

Now, for the matrix 2B, use the scalar multiple property again:
det(2B)=23⋅det(B)=8⋅‌

1

729

=‌

8

729

We now combine these results to find:
|2B(adj(3A))|=det(2B⋅adj(3A))=det(2B)⋅det(adj(3A))
Therefore:
|2B(adj(3A))|=‌

8

729

⋅‌

729

8

=1