Explanation
\begin{aligned}
& \mathrm{A}=
\begin{bmatrix}
2 & 0 & 1 \\
1 & 1 & 0 \\
1 & 0 & 1
\end{bmatrix} \\
& \mathrm{Let~B}_{1}=
\begin{bmatrix}
a \\
b \\
c
\end{bmatrix}\mathrm{B}_{2}=
\begin{bmatrix}
d \\
e \\
f
\end{bmatrix}\mathrm{B}_{3}=
\begin{bmatrix}
g \\
h \\
i
\end{bmatrix} \\
& \mathbf{AB}_{1}={
\begin{bmatrix}
{2a+c} \\
{a+b} \\
{a+c}
\end{bmatrix}}={
\begin{bmatrix}
{1} \\
{0} \\
{0}
\end{bmatrix}}\Rightarrow a=1,b=-1,c=-1 \\
& \mathrm{AB}_{2}=
\begin{bmatrix}
{2d+b} \\
{d+e} \\
{d+f}
\end{bmatrix}=
\begin{bmatrix}
{2} \\
{3} \\
{0}
\end{bmatrix}\Rightarrow d=2,e=1,f=-2 \\
& \mathrm{AB}_{3}=
\begin{bmatrix}
{2g+h} \\
{g+h} \\
{g+i}
\end{bmatrix}=
\begin{bmatrix}
{3} \\
{2} \\
{1}
\end{bmatrix}\Rightarrow g=2,h=0,i=-1
\end{aligned}
\begin{aligned}
& \mathbf{B}=
\begin{bmatrix}
1 & 2 & 2 \\
-1 & 1 & 0 \\
-1 & -2 & -1
\end{bmatrix} \\
& \Rightarrow\left|\mathrm{B}\right|=1\left(-1+0\right)-2\left(1+0\right)+2\left(2+1\right) \\
& =-1-2+6=3 \\
& \therefore\alpha=3,\beta=1+1-1=1 \\
& \therefore\alpha^{3}+\beta^{3}=3^{3}+1^{3}=27
\end{aligned}
