Find x, if

We have:

$\left[\begin{array}{lll}x & -5 & -1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=O$

$\Rightarrow[x+0-2 \quad 0-10+0 \quad 2 x-5-3]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=O$

$\Rightarrow\left[\begin{array}{lll}x-2 & -10 & 2 x-8\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=O$

$\Rightarrow[x(x-2)-40+2 x-8]=O$

$\Rightarrow\left[x^{2}-2 x-40+2 x-8\right]=[0]$

$\Rightarrow\left[x^{2}-48\right]=[0]$

$\therefore x^{2}-48=0$

$\Rightarrow x^{2}=48$

$\Rightarrow x=\pm 4 \sqrt{3}$