Question

Show that x = 2 is a root of the equation

\[\begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 3 & 2x & x + 2\end{vmatrix} = 0\]  and solve it completely.

 

 

Answer 1

\[\text{ Let }∆ = \begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 3 & 2x & x + 2\end{vmatrix}\] 

\[ = \begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 3 - x & 2x + 6 & x + 3\end{vmatrix} \left[\text{ Applying }R_3\text{ to }R_3 - R_1 \right]\] 

\[ = \left( x + 3 \right)\begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 1 & 2 & 1\end{vmatrix} \] 

\[ = \left( x + 3 \right)\begin{vmatrix}x - 2 & 3x - 6 & - x + 2 \\ 2 & - 3x & x - 3 \\ - 1 & 2 & 1\end{vmatrix} \left[\text{ Applying } R_1 \text{ to }R_1 - R_2 \right]\] 

\[ = \left( x + 3 \right)\left( x - 2 \right)\begin{vmatrix}1 & 3 & - 1 \\ 2 & - 3x & x - 3 \\ - 1 & 2 & 1\end{vmatrix} \] 

\[ = \left( x + 3 \right)\left( x - 2 \right)\begin{vmatrix}1 & 3 & 0 \\ 2 & - 3x & x - 1 \\ - 1 & 2 & 0\end{vmatrix} \left[\text{ Applying }C_3 \text{ to }C_3 + C_1 \right]\] 

\[ = \left( x + 3 \right)\left( x - 2 \right)\left( x - 1 \right)\begin{vmatrix}1 & 3 & 0 \\ 2 & - 3x & 1 \\ - 1 & 2 & 0\end{vmatrix} \] 

\[ = \left( x + 3 \right)\left( x - 2 \right)\left( x - 1 \right)\left\{ - 1\begin{vmatrix}1 & 3 \\ - 1 & 2\end{vmatrix} \right\} \left[ \text{ Expanding along }C_3 \right]\] 

\[ = - 5\left( x + 3 \right)\left( x - 2 \right)\left( x - 1 \right)\] 

\[x = 2, - 3, 1\]