If {A_1}, {B_1}, {C_1}, dots are respectively | vedclass

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(A) We know that the cofactor of an element $a_{ij}$ in a determinant $\Delta$ is denoted by $A_{ij}$.For the given determinant $\Delta$,the cofactors

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${a_1}\Delta $

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(A) We know that the cofactor of an element $a_{ij}$ in a determinant $\Delta$ is denoted by $A_{ij}$.For the given determinant $\Delta$,the cofactors are defined as:$B_2 = (-1)^{2+2} \begin{vmatrix} a_1 & c_1 \ a_3 & c_3 \end{vmatrix} = a_1 c_3 - a_3 c_1$$C_2 = (-1)^{2+3} \begin{vmatrix} a_1 & b_1 \ a_3 & b_3 \end{vmatrix} = -(a_1 b_3 - a_3 b_1)$$B_3 = (-1)^{3+2} \begin{vmatrix} a_1 & c_1 \ a_2 & c_2 \end{vmatrix} = -(a_1 c_2 - a_2 c_1)$$C_3 = (-1)^{3+3} \begin{vmatrix} a_1 & b_1 \ a_2 & b_2 \end{vmatrix} = a_1 b_2 - a_2 b_1$Now,consider the determinant $D = \begin{vmatrix} B_2 & C_2 \ B_3 & C_3 \end{vmatrix}$.By the property of adjoint matrices,for any $3 	imes 3$ matrix $M$,the determinant of the cofactor matrix is $\Delta^{n-1}$,where $n=3$. Thus,the determinant of the $2 	imes 2$ minor of the cofactor matrix is $a_1 \Delta$.Substituting the values:$D = (a_1 c_3 - a_3 c_1)(a_1 b_2 - a_2 b_1) - (-(a_1 b_3 - a_3 b_1))(-(a_1 c_2 - a_2 c_1))$Expanding this expression simplifies to $a_1(a_1 b_2 c_3 - a_1 b_3 c_2 - b_1 a_2 c_3 + b_1 a_3 c_2 + c_1 a_2 b_3 - c_1 a_3 b_2) = a_1 \Delta$.Therefore,the correct option is $A$.

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