left| {begin{array}{*{20}{c}}1&1&11&{1 + x}&1 | vedclass

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(D) Let $\Delta = \left| {\begin{array}{*{20}{c}}1&1&1\1&{1 + x}&1\1&1&{1 + y}\end{array}} ight|$.Apply the column operations $C_1 	o C_1 - C_3$

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$xy$

Vedclass · Answer

(D) Let $\Delta = \left| {\begin{array}{*{20}{c}}1&1&1\1&{1 + x}&1\1&1&{1 + y}\end{array}} ight|$.Apply the column operations $C_1 	o C_1 - C_3$ and $C_2 	o C_2 - C_3$:$\Delta = \left| {\begin{array}{*{20}{c}}1-1&1-1&1\1-1&1+x-1&1\1-1&1-1&1+y\end{array}} ight| = \left| {\begin{array}{*{20}{c}}0&0&1\0&x&1\0&0&1+y\end{array}} ight|$.Wait,let us use $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$:$\Delta = \left| {\begin{array}{*{20}{c}}1&0&0\1&x&0\1&0&y\end{array}} ight|$.Expanding along the first row:$\Delta = 1(xy - 0) - 0 + 0 = xy$.Thus,the correct option is $D$.

$\left| {\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}} \right| = $

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