Question
Let \[\Delta = \left| {\begin{array}{*{20}{c}}
{1 + {x_1}{y_1}}&{1 + {x_1}{y_2}}&{1 + {x_1}{y_3}}\\
{1 + {x_2}{y_1}}&{1 + {x_2}{y_2}}&{1 + {x_2}{y_3}}\\
{1 + {x_3}{y_1}}&{1 + {x_3}{y_2}}&{1 + {x_3}{y_3}}
\end{array}} \right|,\] then value of $$\Delta $$ is
A.
$${x_1}{x_2}{x_3} + {y_1}{y_2}{y_3}$$
B.
$${x_1}{x_2}{x_3}{y_1}{y_2}{y_3}$$
C.
$${x_2}{x_3}{y_2}{y_3} + {x_3}{x_1}{y_3}{y_1} + {x_1}{x_2}{y_1}{y_2}$$
D.
$$0$$
Answer :
$$0$$
Solution :
We can write $$\Delta = {\Delta _1} + {y_1}{\Delta _2},{\text{ where}}$$
\[\begin{array}{l}
{\Delta _1} = \left| {\begin{array}{*{20}{c}}
1&{1 + {x_1}{y_2}}&{1 + {x_1}{y_3}}\\
1&{1 + {x_2}{y_2}}&{1 + {x_2}{y_3}}\\
1&{1 + {x_3}{y_2}}&{1 + {x_3}{y_3}}
\end{array}} \right|{\rm{ and}}\\
{\Delta _2} = \left| {\begin{array}{*{20}{c}}
{{x_1}}&{1 + {x_1}{y_2}}&{1 + {x_1}{y_3}}\\
{{x_2}}&{1 + {x_2}{y_2}}&{1 + {x_2}{y_3}}\\
{{x_3}}&{1 + {x_3}{y_2}}&{1 + {x_3}{y_3}}
\end{array}} \right|
\end{array}\]
$$\eqalign{
& {\text{In }}{\Delta _1},{\text{use }}{C_2} \to {C_2} - {C_1}{\text{ and }}{C_3} \to {C_3} - {C_1} \cr
& {\text{so that,}} \cr} $$
\[{\Delta _1} = \left| {\begin{array}{*{20}{c}}
1&{{x_1}{y_2}}&{{x_1}{y_3}}\\
1&{{x_2}{y_2}}&{{x_2}{y_3}}\\
1&{{x_3}{y_2}}&{{x_3}{y_3}}
\end{array}} \right| = 0\]
$$\eqalign{
& \left[ {\because {C_2}{\text{ and }}{C_3}{\text{ are proportional}}} \right] \cr
& {\text{In }}{\Delta _2},{\text{us }}{C_2} \to {C_2} - {y_2}{C_1} \cr
& {\text{and }}{C_3} \to {C_3} - {y_3}{C_1}{\text{ to get}} \cr} $$
\[{\Delta _2} = \left| {\begin{array}{*{20}{c}}
{{x_1}}&1&1 \\
{{x_2}}&1&1 \\
{{x_3}}&1&1
\end{array}} \right| = 0\left[ {\because {C_2}{\text{ and }}{C_3}{\text{ are identical}}} \right]\]
$$\therefore \Delta = 0$$