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Question
Show that x = 2 is a root of the equation
Solution
\[\text{ Let }∆ = \begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 3 & 2x & x + 2\end{vmatrix}\]
\[ = \begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 3 - x & 2x + 6 & x + 3\end{vmatrix} \left[\text{ Applying }R_3\text{ to }R_3 - R_1 \right]\]
\[ = \left( x + 3 \right)\begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 1 & 2 & 1\end{vmatrix} \]
\[ = \left( x + 3 \right)\begin{vmatrix}x - 2 & 3x - 6 & - x + 2 \\ 2 & - 3x & x - 3 \\ - 1 & 2 & 1\end{vmatrix} \left[\text{ Applying } R_1 \text{ to }R_1 - R_2 \right]\]
\[ = \left( x + 3 \right)\left( x - 2 \right)\begin{vmatrix}1 & 3 & - 1 \\ 2 & - 3x & x - 3 \\ - 1 & 2 & 1\end{vmatrix} \]
\[ = \left( x + 3 \right)\left( x - 2 \right)\begin{vmatrix}1 & 3 & 0 \\ 2 & - 3x & x - 1 \\ - 1 & 2 & 0\end{vmatrix} \left[\text{ Applying }C_3 \text{ to }C_3 + C_1 \right]\]
\[ = \left( x + 3 \right)\left( x - 2 \right)\left( x - 1 \right)\begin{vmatrix}1 & 3 & 0 \\ 2 & - 3x & 1 \\ - 1 & 2 & 0\end{vmatrix} \]
\[ = \left( x + 3 \right)\left( x - 2 \right)\left( x - 1 \right)\left\{ - 1\begin{vmatrix}1 & 3 \\ - 1 & 2\end{vmatrix} \right\} \left[ \text{ Expanding along }C_3 \right]\]
\[ = - 5\left( x + 3 \right)\left( x - 2 \right)\left( x - 1 \right)\]
\[x = 2, - 3, 1\]
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