સમીકરણ left| {,begin{array}{*{20}{c}}1&4& | vedclass

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Question

(b) $\left| {\,\begin{array}{*{20}{c}}1&4&{20}\1&{ - 2}&5\1&{2x}&{5{x^2}}\end{array}\,\,} ight|\, = 0$

$ \Rightarrow $

Vedclass · Accepted Answer

$ - 1,\,2$

Vedclass · Answer

(b) $\left| {\,\begin{array}{*{20}{c}}1&4&{20}\1&{ - 2}&5\1&{2x}&{5{x^2}}\end{array}\,\,} ight|\, = 0$

$ \Rightarrow $ $\left| {\,\begin{array}{*{20}{c}}0&6&{15}\0&{ - 2 - 2x}&{5(1 - {x^2})}\1&{2x}&{5{x^2}}\end{array}\,} ight|\, = 0$ $\left( \begin{array}{l}{R_1} 	o {R_1} - {R_2}\{R_2} 	o {R_2} - {R_3}\end{array} ight)$

$ \Rightarrow $$3\,.\,2\,.\,5.\,\left| {\,\begin{array}{*{20}{c}}0&1&1\0&{ - (1 + x)}&{1 - {x^2}}\1&x&{{x^2}}\end{array}\,} ight| = 0$

$ \Rightarrow $ $(1 + x)\,\left| {\,\begin{array}{*{20}{c}}0&1&1\0&{ - 1}&{1 - x}\1&x&{{x^2}}\end{array}\,} ight|\, = 0$

$ \Rightarrow $ $x + 1 = 0$ or $x - 2 = 0$ $ \Rightarrow $ $x = - 1,\,2$.

Trick: Obviously by inspection, $x = - 1,\,2$ satisfy the equation.

At $x = - 1,\,$ $\left| {\,\begin{array}{*{20}{c}}1&4&{20}\1&{ - 2}&5\1&{ - 2}&5\end{array}\,} ight|\, = 0$ as ${R_2} \equiv {R_3}$

At $x = 2$, $\left| {\,\begin{array}{*{20}{c}}1&4&{20}\1&{ - 2}&5\1&4&{20}\end{array}\,} ight| = 0$ as ${R_1} \equiv {R_3}$.

સમીકરણ $\left| {\,\begin{array}{*{20}{c}}1&4&{20}\\1&{ - 2}&5\\1&{2x}&{5{x^2}}\end{array}\,} \right| = 0$ ના બીજ મેળવો.

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