The roots of the equation left| {,begin{array | vedclass

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Question

(b) $\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\1&{1 + x}&1\1&1&{1 + x}\end{array}\,} ight|\, = \,0$

$ \Rightarrow $$

Vedclass · Accepted Answer

$0, 0, -3$

Vedclass · Answer

(b) $\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\1&{1 + x}&1\1&1&{1 + x}\end{array}\,} ight|\, = \,0$

$ \Rightarrow $$\left| {\,\begin{array}{*{20}{c}}
{3 + x}&0&1\
{3 + x}&x&1\
{3 + x}&{ - x}&{1 + x}
\end{array}\,} ight|\, = \,0$, $\left( \begin{array}{l}{C_1} 	o {C_1} + {C_2} + {C_3}\{C_2} 	o {C_2} - {C_3}\end{array} ight)$

$ \Rightarrow $ $(x + 3)\,\left| {\,\begin{array}{*{20}{c}}1&0&1\1&x&1\1&{ - x}&{1 + x}\end{array}\,} ight|\, = 0$

$ \Rightarrow $ $(x + 3)\,\left| {\,\begin{array}{*{20}{c}}1&0&1\0&x&0\0&{ - x}&x\end{array}\,} ight|\, = 0$, $\left( \begin{array}{l}{R_2} 	o {R_2} - {R_1}\{R_3} 	o {R_3} - {R_1}\end{array} ight)$

$ \Rightarrow $ $(x + 3){x^2} = 0 \Rightarrow x = 0,\,0,\, - 3$.

Trick : Obviously the equation is of degree three, therefore it must have three solutions.

So check for option $ (b).$

The roots of the equation $\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + x}&1\\1&1&{1 + x}\end{array}\,} \right| = 0$ are

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