Question

Find $x$, if $\left[\begin{array}{ccc}x& -5& -1\end{array}\right]\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right]\left[\begin{array}{c}x\\ 4\\ 1\end{array}\right]=0$

Answer 1

We have :
$\left[\begin{array}{ccc}x& -5& -1\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 2\end{array}\begin{array}{c}0\\ 2\\ 0\end{array}\begin{array}{c}2\\ 1\\ 3\end{array}\right]\left[\begin{array}{c}x\\ 4\\ 1\end{array}\right]=0$
$\Rightarrow \left[\begin{array}{ccc}x+0-2& 0-10+0& 2x-5-3\end{array}\right]\left[\begin{array}{c}x\\ 4\\ 1\end{array}\right]=0$
$\Rightarrow \left[\begin{array}{ccc}x-2& -10& 2x-8\end{array}\right]\left[\begin{array}{c}x\\ 4\\ 1\end{array}\right]=0$
$\Rightarrow [x(x-2)-40+2x-8]=0$
$\Rightarrow [x^2-2x-40+2x-8]=\left[0\right]$
$\Rightarrow [x^2-48]=\left[0\right]$
$\therefore x^2-48=0$
$\Rightarrow x^2=48$
$\Rightarrow x=\pm 4\sqrt{3}$