Question

Show that x = 2 is a root of the equation

$\left|\begin{array}{ccc}x& -6& -1\\ 2& -3x& x-3\\ -3& 2x& x+2\end{array}\right|=0$ and solve it completely.

Answer 1

$$\begin{gathered} \mathrm{Let}∆=\left|\begin{array}{ccc}x& -6& -1\\ 2& -3x& x-3\\ -3& 2x& x+2\end{array}\right| \\ =\left|\begin{array}{ccc}x& -6& -1\\ 2& -3x& x-3\\ -3-x& 2x+6& x+3\end{array}\right|\left[\mathrm{Applying}{R}_3\to R_3-R_1\right] \\ =\left(x+3\right)\left|\begin{array}{ccc}x& -6& -1\\ 2& -3x& x-3\\ -1& 2& 1\end{array}\right| \\ =\left(x+3\right)\left|\begin{array}{ccc}x-2& 3x-6& -x+2\\ 2& -3x& x-3\\ -1& 2& 1\end{array}\right|\left[\mathrm{Applying}{R}_1\to R_1-R_2\right] \\ =\left(x+3\right)\left(x-2\right)\left|\begin{array}{ccc}1& 3& -1\\ 2& -3x& x-3\\ -1& 2& 1\end{array}\right| \\ =\left(x+3\right)\left(x-2\right)\left|\begin{array}{ccc}1& 3& 0\\ 2& -3x& x-1\\ -1& 2& 0\end{array}\right|\left[\mathrm{Applying}{C}_3\to C_3+C_1\right] \\ =\left(x+3\right)\left(x-2\right)\left(x-1\right)\left|\begin{array}{ccc}1& 3& 0\\ 2& -3x& 1\\ -1& 2& 0\end{array}\right| \\ =\left(x+3\right)\left(x-2\right)\left(x-1\right)\left\{-1\left|\begin{array}{cc}1& 3\\ -1& 2\end{array}\right|\right\}\left[\mathrm{Expanding}\mathrm{along}{\mathrm{C}}_3\right] \\ =-5\left(x+3\right)\left(x-2\right)\left(x-1\right) \\ x=2,-3,1 \end{gathered}$$