Match List-I with List-II Let A be any invertible square matrix. Then…

The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

(A) \(A - A^T\)

This matches (II) Skew-symmetric:

    \((A - A^T)^T = A^T - A = -(A - A^T)\)

(B) \(AA^T\)

This matches (III) Symmetric, by the reversal law for transposes:

    \((AA^T)^T = (A^T)^T A^T = AA^T\)

(C) \(\det (A^{-1})\)

This matches (IV) \([\det(A)]^{-1}\). Since \(AA^{-1} = I\):

    \(\det(A)\,\det(A^{-1}) = \det(I) = 1\)

    \( \displaystyle \det(A^{-1}) = \frac{1}{\det(A)} = [\det(A)]^{-1} \)

(D) \(\mathrm{adj}A\)

This matches (I) \(|A|A^{-1}\). From \(A\,(\mathrm{adj}A) = |A|\,I\), multiplying on the left by \(A^{-1}\):

    \(\mathrm{adj}A = |A|\,A^{-1}\)

So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).