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(B) Let $\Delta'$ be the given determinant of co-factors.We know that the product of a determinant and its co-factor matrix determinant is given by $\

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$\Delta^2$

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(B) Let $\Delta'$ be the given determinant of co-factors.We know that the product of a determinant and its co-factor matrix determinant is given by $\Delta \cdot \Delta' = \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} \cdot \begin{vmatrix} A_1 & B_1 & C_1 \ A_2 & B_2 & C_2 \ A_3 & B_3 & C_3 \end{vmatrix}$.By the property of multiplication of determinants,this equals $\begin{vmatrix} a_1A_1 + b_1B_1 + c_1C_1 & a_1A_2 + b_1B_2 + c_1C_2 & a_1A_3 + b_1B_3 + c_1C_3 \ a_2A_1 + b_2B_1 + c_2C_1 & a_2A_2 + b_2B_2 + c_2C_2 & a_2A_3 + b_2B_3 + c_2C_3 \ a_3A_1 + b_3B_1 + c_3C_1 & a_3A_2 + b_3B_2 + c_3C_2 & a_3A_3 + b_3B_3 + c_3C_3 \end{vmatrix}$.Using the property that the sum of the product of elements of a row with their corresponding co-factors is $\Delta$,and the sum of the product of elements of a row with co-factors of another row is $0$,we get:$\Delta \cdot \Delta' = \begin{vmatrix} \Delta & 0 & 0 \ 0 & \Delta & 0 \ 0 & 0 & \Delta \end{vmatrix}$.Evaluating this diagonal determinant,we get $\Delta \cdot \Delta' = \Delta^3$.Therefore,$\Delta' = \Delta^2$ (assuming $\Delta 
eq 0$).

If $\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ and $A_1, B_1, C_1$ denote the co-factors of $a_1, b_1, c_1$ respectively,then the value of the determinant $\begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix}$ is

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