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# If $\Delta=\begin{vmatrix} 1+x_1y_1&1+x_1y_2&1+x_1y_3\\1+x_2y_1&1+x_2y_2&1+x_2y_3\\1+x_3y_1&1+x_3y_2&1+x_3y_3\end{vmatrix}$ then $\Delta$ is

$\begin{array}{1 1}(A)\;x_1x_2x_3+y_1y_2y_3\\(B)\;x_1x_2x_3y_1y_2y_3\\(C)\;x_2x_3y_2y_3+x_3x_1y_3y_1+x_1x_2y_1y_2\\(D)\;\text{None of these}\end{array}$

Can you answer this question?

$\Delta=\Delta_1+y_1\Delta_2$
Where
$\Delta_1=\begin{vmatrix}1& 1+x_1y_2&1+x_1y_3\\1&1+x_2y_2&1+x_2y_3\\1&1+x_3y_2&1+x_3y_3\end{vmatrix}$
$\Delta_2=\begin{vmatrix}x_1& 1+x_1y_2&1+x_1y_3\\x_2&1+x_2y_2&1+x_2y_3\\x_3&1+x_3y_2&1+x_3y_3\end{vmatrix}$
Apply $C_2-C_1,C_3-c_1$ in $\Delta_1$ we get
$\Delta_1=\begin{vmatrix}1& x_1y_2&x_1y_3\\1&x_2y_2&x_2y_3\\1&x_3y_2&x_3y_3\end{vmatrix}=0$
Operating $C_2-y_2C_1$ & $C_3-y_3C_1$ we get
$\Delta_1=\begin{vmatrix}x_1& 1&1\\x_2&1&1\\x_3&1&1\end{vmatrix}=0$
Hence $\Delta=0$
Hence (D) is the correct answer.
answered Apr 22, 2014