Solve the matrix equations:

(i) $\left[\begin{array}{ll}x & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -2 & -3\end{array}\right]\left[\begin{array}{l}x \\ 5\end{array}\right]=0$

$\Rightarrow\left[\begin{array}{ll}x-2 & 0-3\end{array}\right]\left[\begin{array}{l}x \\ 5\end{array}\right]=0$

$\Rightarrow\left[\begin{array}{ll}x-2 & -3\end{array}\right]\left[\begin{array}{l}x \\ 5\end{array}\right]=0$

$\Rightarrow\left[x^{2}-2 x-15\right]=0$

$\Rightarrow x^{2}-2 x-15=0$

$\Rightarrow x^{2}-5 x+3 x-15=0$

$\Rightarrow x(x-5)+3(x-5)=0$

$\Rightarrow(x-5)(x+3)=0$

$\Rightarrow x-5=0$ or $x+3=0$

$\Rightarrow x=5$ or $x=-3$

(ii) $\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0$

$\Rightarrow\left[\begin{array}{lll}1+4+1 & 2+0+0 & 0+2+2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0$

$\Rightarrow\left[\begin{array}{lll}6 & 2 & 4\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=0$

$\Rightarrow[0+4+4 x]=0$

$\Rightarrow 4+4 x=0$

$\Rightarrow 4 x=-4$

$\therefore x=\frac{-4}{4}=-1$

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

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

$\Rightarrow\left[\begin{array}{lll}x-2 & -10 & 2 x-8\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$

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

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

$\Rightarrow x^{2}=48$

$\Rightarrow x=\pm \sqrt{48}$

$($ iv $)\left[\begin{array}{ll}2 x & 3\end{array}\right]\left[\begin{array}{cc}1 & 2 \\ -3 & 0\end{array}\right]\left[\begin{array}{l}x \\ 8\end{array}\right]=0$

$\Rightarrow\left[\begin{array}{ll}2 x-9 & 4 x\end{array}\right]\left[\begin{array}{l}x \\ 8\end{array}\right]=0$

$\Rightarrow[x(2 x-9)+32 x]=0$

$\Rightarrow\left[2 x^{2}-9 x+32 x\right]=0$

$\Rightarrow\left[2 x^{2}+23 x\right]=0$

$\Rightarrow 2 x^{2}+23 x=0$

$\Rightarrow x(2 x+23)=0$

$\Rightarrow x=0$ or $x=-\frac{23}{2}$

$\therefore x=0$ or $x=-\frac{23}{2}$