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(B) Let the determinant be $\Delta(x)$.To find $A_1$,we differentiate $\Delta(x)$ with respect to $x$ and evaluate at $x = 0$,i.e.,$A_1 = \Delta'(0)$.

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$0$

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(B) Let the determinant be $\Delta(x)$.To find $A_1$,we differentiate $\Delta(x)$ with respect to $x$ and evaluate at $x = 0$,i.e.,$A_1 = \Delta'(0)$.Using the property of differentiation of determinants,$\Delta'(x)$ is the sum of three determinants where each row is differentiated one by one.$\Delta'(x) = \left| \begin{array}{ccc} a & b & c \ 1+a_1x & 1+b_1x & 1+c_1x \ 1+a_2x & 1+b_2x & 1+c_2x \end{array} ight| + \left| \begin{array}{ccc} 1+ax & 1+bx & 1+cx \ a_1 & b_1 & c_1 \ 1+a_2x & 1+b_2x & 1+c_2x \end{array} ight| + \left| \begin{array}{ccc} 1+ax & 1+bx & 1+cx \ 1+a_1x & 1+b_1x & 1+c_1x \ a_2 & b_2 & c_2 \end{array} ight|$.Evaluating at $x = 0$:$\Delta'(0) = \left| \begin{array}{ccc} a & b & c \ 1 & 1 & 1 \ 1 & 1 & 1 \end{array} ight| + \left| \begin{array}{ccc} 1 & 1 & 1 \ a_1 & b_1 & c_1 \ 1 & 1 & 1 \end{array} ight| + \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & 1 & 1 \ a_2 & b_2 & c_2 \end{array} ight|$.Since each determinant has two identical rows,the value of each determinant is $0$.Therefore,$A_1 = 0 + 0 + 0 = 0$.

If $\left| \begin{array}{ccc} 1 + ax & 1 + bx & 1 + cx \\ 1 + a_1x & 1 + b_1x & 1 + c_1x \\ 1 + a_2x & 1 + b_2x & 1 + c_2x \end{array} \right| = A_0 + A_1x + A_2x^2 + A_3x^3$,then $A_1$ is equal to:

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