$\begin{aligned} & \text { (c) }\left[\begin{array}{lll}x & 4 & -1\end{array}\right]\left[\begin{array}{lll}2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4\end{array}\right]\left[\begin{array}{c}x \\ 4 \\ -1\end{array}\right]=0 \\ & \Rightarrow\left[\begin{array}{lll}2 x+4 & x-2 & 4\end{array}\right]\left[\begin{array}{c}x \\ 4 \\ -1\end{array}\right]=0 \\ & \Rightarrow x(2 x+4)+4(x-2)-4=0 \\ & \Rightarrow 2 x^2+4 x+4 x-8-4=0 \\ & \Rightarrow 2 x^2+8 x-12=0 \Rightarrow x^2+4 x-6=0 \\ & \therefore x=\frac{-4 \pm \sqrt{16+24}}{2}=\frac{-4 \pm 2 \sqrt{10}}{2} \\ & \Rightarrow x=-2 \pm \sqrt{10}\end{aligned}$