left| {begin{array}{*{20}{c}}{x + 1}&{x + 2}& | vedclass

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(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\{x + 3}&{x + 5}&{x + 8}\{x + 7}&{x + 10}&{x + 14}\end{array}} ight|$.Appl

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\{x + 3}&{x + 5}&{x + 8}\{x + 7}&{x + 10}&{x + 14}\end{array}} ight|$.Applying $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_2$:$\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{1}&{2}\{x + 3}&{2}&{3}\{x + 7}&{3}&{4}\end{array}} ight|$.Applying $C_3 	o C_3 - C_2$:$\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{1}&{1}\{x + 3}&{2}&{1}\{x + 7}&{3}&{1}\end{array}} ight|$.Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_2$:$\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{1}&{1}\{2}&{1}&{0}\{4}&{1}&{0}\end{array}} ight|$.Expanding along $C_3$:$\Delta = 1 	imes (2 - 4) = -2$.Thus,the value of the determinant is $-2$.

$\left| {\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}} \right| = $

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