Show that x = 2 is a root of the equation

Question:

Show that $x=2$ is a root of the equation

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

Solution:

Let $\Delta=\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & x+2\end{array}\right|$

$=\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3-x & 2 x+6 & x+3\end{array}\right| \quad\left[\right.$ Applying $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$

$=(x+3)\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -1 & 2 & 1\end{array}\right|$

$=(x+3)\left|\begin{array}{ccc}x-2 & 3 x-6 & -x+2 \\ 2 & -3 x & x-3 \\ -1 & 2 & 1\end{array}\right| \quad$ [Applying $R_{1} \rightarrow R_{1}-R_{2}$ ]

$=(x+3)(x-2)\left|\begin{array}{ccc}1 & 3 & -1 \\ 2 & -3 x & x-3 \\ -1 & 2 & 1\end{array}\right|$

$=(x+3)(x-2)\left|\begin{array}{ccc}1 & 3 & 0 \\ 2 & -3 x & x-1 \\ -1 & 2 & 0\end{array}\right| \quad$ [Applying $\left.C_{3} \rightarrow C_{3}+C_{1}\right]$

$=(x+3)(x-2)(x-1)\left|\begin{array}{ccc}1 & 3 & 0 \\ 2 & -3 x & 1 \\ -1 & 2 & 0\end{array}\right|$

$=(x+3)(x-2)(x-1)\left\{-1\left|\begin{array}{cc}1 & 3 \\ -1 & 2\end{array}\right|\right\} \quad$ [Expanding along $\left.\mathrm{C}_{3}\right]$

$=-5(x+3)(x-2)(x-1)$

$x=2,-3,1$