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

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(A) Let $\Delta = \left| {\begin{array}{*{20}{c}}{1 + x}&1&1\1&{1 + y}&1\1&1&{1 + z}\end{array}} ight|$.Taking $x, y, z$ common from $C_1, C_2, C_

Vedclass · Accepted Answer

$xyz\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$

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(A) Let $\Delta = \left| {\begin{array}{*{20}{c}}{1 + x}&1&1\1&{1 + y}&1\1&1&{1 + z}\end{array}} ight|$.Taking $x, y, z$ common from $C_1, C_2, C_3$ respectively:$\Delta = xyz \left| {\begin{array}{*{20}{c}}{1 + \frac{1}{x}}&{\frac{1}{y}}&{\frac{1}{z}}\1&{1 + \frac{1}{y}}&{\frac{1}{z}}\1&{\frac{1}{y}}&{1 + \frac{1}{z}}\end{array}} ight|$ (This is not the standard approach,let's use row operations).Alternative approach: Factor out $x, y, z$ from the columns:$\Delta = xyz \left| {\begin{array}{*{20}{c}}{\frac{1}{x} + 1}&{\frac{1}{y}}&{\frac{1}{z}}\1&{1 + \frac{1}{y}}&{\frac{1}{z}}\1&{\frac{1}{y}}&{1 + \frac{1}{z}}\end{array}} ight|$Applying $R_1 	o R_1 + R_2 + R_3$ is not ideal here. Let's use $C_1 	o C_1/x, C_2 	o C_2/y, C_3 	o C_3/z$ is not correct.Correct method: $\Delta = xyz \left| {\begin{array}{*{20}{c}}{\frac{1}{x}+1}&{\frac{1}{y}}&{\frac{1}{z}}\ \frac{1}{x} & \frac{1}{y}+1 & \frac{1}{z} \ \frac{1}{x} & \frac{1}{y} & \frac{1}{z}+1 \end{array}} ight|$.Applying $C_1 	o C_1 + C_2 + C_3$:$\Delta = xyz (1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \left| {\begin{array}{*{20}{c}}1&{\frac{1}{y}}&{\frac{1}{z}}\1&{1 + \frac{1}{y}}&{\frac{1}{z}}\1&{\frac{1}{y}}&{1 + \frac{1}{z}}\end{array}} ight|$.Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\Delta = xyz (1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \left| {\begin{array}{*{20}{c}}1&{\frac{1}{y}}&{\frac{1}{z}}\0&1&0\0&0&1\end{array}} ight| = xyz (1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z})$.Trick: Put $x = 1, y = 2, z = 3$:$\Delta = \left| {\begin{array}{*{20}{c}}2&1&1\1&3&1\1&1&4\end{array}} ight| = 2(12-1) - 1(4-1) + 1(1-3) = 2(11) - 3 - 2 = 22 - 5 = 17$.Option $(a): 1 	imes 2 	imes 3 (1 + 1 + 0.5 + 0.333) = 6 (2.833) = 17$.

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

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