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(C) For a non-singular matrix $A$ of order $n$,$|	ext{adj}(A)| = |A|^{n-1}$. Here $n=3$ and $|A|=a$,so $|	ext{adj}(A)| = a^{3-1} = a^2$. Thus,$(A)-(

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$A$-$II$,$B$-$I$,$C$-$IV$,$D$-$V$

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(C) For a non-singular matrix $A$ of order $n$,$|	ext{adj}(A)| = |A|^{n-1}$. Here $n=3$ and $|A|=a$,so $|	ext{adj}(A)| = a^{3-1} = a^2$. Thus,$(A)-(II)$.$(B)$ Given $AB=O$ and $A$ is non-singular $(|A| 
eq 0)$. Multiplying by $A^{-1}$ on the left,$A^{-1}(AB) = A^{-1}O \implies (A^{-1}A)B = O \implies IB = O \implies B=O$. Thus,$B$ is a null matrix. $(B)-(I)$.$(C)$ The determinant $\Delta = \begin{vmatrix} 1 & x & x^2 \ \cos(a-b)y & \cos ay & \cos(a+b)y \ \sin(a-b)y & \sin ay & \sin(a+b)y \end{vmatrix}$. Using $C_1 	o C_1 + C_3$,we get terms involving $2\cos(ay)\cos(by)$ and $2\sin(ay)\cos(by)$. Simplifying,the determinant is independent of $a$. Thus,$(C)-(IV)$.$(D)$ $B = A - A^T$. Then $B^T = (A - A^T)^T = A^T - A = -(A - A^T) = -B$. Thus,$B$ is a skew-symmetric matrix. For a skew-symmetric matrix of odd order $3$,the determinant is $0$. Thus,$(D)-(V)$.

List-$I$	List-$II$
$(A)$ If $A$ is a non-singular matrix of order $3$ and $\|A\|=a$,then $\|\text{adj}(A)\|=$	$(I)$ null matrix
$(B)$ $A$ is a non-singular matrix of order $3$ and $B$ is any matrix of order $3$ such that $AB=O$,then $B$ is	$(II)$ $a^2$
$(C)$ $\begin{vmatrix} 1 & x & x^2 \\ \cos(a-b)y & \cos ay & \cos(a+b)y \\ \sin(a-b)y & \sin ay & \sin(a+b)y \end{vmatrix}$ does not depend on	$(III)$ $b$
$(D)$ $A$ is a square matrix of order $3$ and $B=A-A^T$,then $B$ is	$(IV)$ $a$
	$(V)$ $0$

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