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

$ \Rightarrow $ $\left| {\,\begin{array}{*{20}{c}}0&6&{15}\\0&{ – 2 – 2x}&{5(1 – {x^2})}\\1&{2x}&{5{x^2}}\end{array}\,} \right|\, = 0$

$\left( \begin{array}{l}{R_1} \to {R_1} – {R_2}\\{R_2} \to {R_2} – {R_3}\end{array} \right)$

$ \Rightarrow $ $3\,.\,2\,.\,5.\,\left| {\,\begin{array}{*{20}{c}}0&1&1\\0&{ – (1 + x)}&{1 – {x^2}}\\1&x&{{x^2}}\end{array}\,} \right| = 0$

$ \Rightarrow $ $(1 + x)\,\left| {\,\begin{array}{*{20}{c}}0&1&1\\0&{ – 1}&{1 – x}\\1&x&{{x^2}}\end{array}\,} \right|\, = 0$

$ \Rightarrow $ $x + 1 = 0$ or $x – 2 = 0$

$ \Rightarrow $ $x = – 1,\,2$. 

ट्रिक: जाँच द्वारा स्पष्ट है $x =  – 1,\,2$ समीकरण को सन्तुष्ट करते हैं।  $x =  – 1$ पर,

$\left| {\,\begin{array}{*{20}{c}}1&4&{20}\\1&{ – 2}&5\\1&{ – 2}&5\end{array}\,} \right|\, = 0$,  (चूकि ${R_2} \equiv {R_3}$)

$x = 2$ पर, $\left| {\,\begin{array}{*{20}{c}}1&4&{20}\\1&{ – 2}&5\\1&4&{20}\end{array}\,} \right| = 0$,   (चूकि ${R_1} \equiv {R_3}$)