Match List-I with List-II Let A be any invertible square matrix. Then | List-I | List-II | | --- | --- | | (A) A - A^T | (I) vert Avert A^-1 | | (B)…

$$(A - A^T)^T = A^T - (A^T)^T$$(A−AT)T=AT−(AT)T

$$= A^T - A$$=AT−A

$$= -(A - A^T)$$=−(A−AT)

Since the transpose equals the negative of itself, $A - A^T$A−AT is skew-symmetric.

$$(A) \rightarrow \text{(II)}$$(A)→(II)

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$$(AA^T)^T = (A^T)^T \cdot A^T$$(AAT)T=(AT)T⋅AT

$$= A \cdot A^T$$=A⋅AT

$$= AA^T$$=AAT

Since the transpose equals itself, $AA^T$AAT is symmetric.

$$(B) \rightarrow \text{(III)}$$(B)→(III)

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Since $A \cdot A^{-1} = I$A⋅A−1=I, taking determinant on both sides:

$$\det(A) \cdot \det(A^{-1}) = \det(I) = 1$$det(A)⋅det(A−1)=det(I)=1

$$\det(A^{-1}) = \dfrac{1}{\det(A)}$$det(A−1)=det(A)1​

$$= [\det(A)]^{-1}$$=[det(A)]−1

$$(C) \rightarrow \text{(IV)}$$(C)→(IV)

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From the standard identity:

$$A^{-1} = \dfrac{\text{adj}(A)}{\det(A)}$$A−1=det(A)adj(A)​

$$\text{adj}(A) = \det(A) \cdot A^{-1}$$adj(A)=det(A)⋅A−1

$$= |A|\,A^{-1}$$=∣A∣A−1

$$(D) \rightarrow \text{(I)}$$(D)→(I)

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List-I	List-II
(A) $A - A^T$A−AT	(II) Skew-symmetric
(B) $AA^T$AAT	(III) Symmetric
(C) $\det(A^{-1})$det(A−1)	(IV) $[\det(A)]^{-1}$[det(A)]−1
(D) $\text{adj}(A)$adj(A)	(I) $\|A\|A^{-1}$∥A∥A−1