If operatorname{det}(AB)=(operatorname{det} A | vedclass

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(C) For any square matrix $A$ of order $n 	imes n$,the property of the adjoint matrix is given by $\operatorname{adj}(A) \cdot A = \operatorname{det}

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$(\operatorname{det}(A))^2$

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(C) For any square matrix $A$ of order $n 	imes n$,the property of the adjoint matrix is given by $\operatorname{adj}(A) \cdot A = \operatorname{det}(A) \cdot I_n$. Taking the determinant on both sides,we get $\operatorname{det}(\operatorname{adj}(A) \cdot A) = \operatorname{det}(\operatorname{det}(A) \cdot I_n)$. Using the property $\operatorname{det}(AB) = \operatorname{det}(A) \operatorname{det}(B)$,we have $\operatorname{det}(\operatorname{adj}(A)) \cdot \operatorname{det}(A) = (\operatorname{det}(A))^n \cdot \operatorname{det}(I_n)$. Since $\operatorname{det}(I_n) = 1$,this simplifies to $\operatorname{det}(\operatorname{adj}(A)) \cdot \operatorname{det}(A) = (\operatorname{det}(A))^n$. For a matrix of order $n=3$,we have $\operatorname{det}(\operatorname{adj}(A)) \cdot \operatorname{det}(A) = (\operatorname{det}(A))^3$. Dividing both sides by $\operatorname{det}(A)$ (since $A$ is non-singular,$\operatorname{det}(A) 
eq 0$),we get $\operatorname{det}(\operatorname{adj}(A)) = (\operatorname{det}(A))^2$.

If $\operatorname{det}(AB)=(\operatorname{det} A)(\operatorname{det} B)$ and $A$ is a non-singular matrix of order $3 \times 3$,then $\operatorname{det}(\operatorname{adj} A)=$

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