Question

Using the properties of determinants, solve the following for x:

$\left|\begin{array}{ccc}x+2& x+6& x-1\\ x+6& x-1& x+2\\ x-1& x+2& x+6\end{array}\right|=0$

Answer 1

$Let\Delta =\left|\begin{array}{ccc}x+2& x+6& x-1\\ x+6& x-1& x+2\\ x-1& x+2& x+6\end{array}\right|Applying{C}_2\to C_2-C_{1}and{C}_3\to C_3-C_1$
$\Delta =\left|\begin{array}{ccc}x+2& 4& -3\\ x+6& -7& -4\\ x-1& 3& 7\end{array}\right|Applying{R}_2\to R_2-R_{1}and{R}_3\to R_3-R_1$
$\Delta =\left|\begin{array}{ccc}x+2& 4& -3\\ 4& -11& -1\\ -3& -1& 10\end{array}\right|Applying{R}_2\to R_2+R_3$
$\Delta =\left|\begin{array}{ccc}x+2& 4& -3\\ 1& -12& 9\\ -3& -1& 10\end{array}\right|$

$Applying{R}_3\to R_3+\left(3\right)R_2\Delta =\left|\begin{array}{ccc}x+2& 4& -3\\ 1& -12& 9\\ 0& -37& 37\end{array}\right|$
$Expandingalong{C}_1\Delta =(x+2)\left|\begin{array}{cc}-12& 9\\ -37& 37\end{array}\right|-1\left|\begin{array}{cc}4& -3\\ -37& 37\end{array}\right|$
$\Delta =(x+2)(-444)+333-1(148-111)\Delta =(x+2)(-111)-1\left(37\right)\therefore \Delta =0=-111x-259\therefore x=\frac{259}{111}=-\frac{7}{3}$