Answer (28)

Solution

$A=\left[\begin{array}{lll}2 & 0 & 1 \ 1 & 1 & 0 \ 1 & 0 & 1\end{array}\right] \quad \mathrm{B}=\left[\mathrm{B}{1}, \mathrm{~B}{2}, \mathrm{~B}_{3}\right]$

$\mathrm{B}{1}=\left[\begin{array}{l}\mathrm{x}{1} \ \mathrm{y}{1} \ \mathrm{z}{1}\end{array}\right], \quad \mathrm{B}{2}=\left[\begin{array}{l}\mathrm{x}{2} \ \mathrm{y}{2} \ \mathrm{z}{2}\end{array}\right], \quad \mathrm{B}{3}=\left[\begin{array}{l}\mathrm{x}{3} \ \mathrm{y}{3} \ \mathrm{z}{3}\end{array}\right]$

$\mathrm{AB}{1}=\left[\begin{array}{lll}2 & 0 & 1 \ 1 & 1 & 0 \ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x{1} \ y_{1} \ z_{1}\end{array}\right]=\left[\begin{array}{l}1 \ 0 \ 0\end{array}\right]$

$\mathrm{x}{1}=1, \mathrm{y}{1}=-1, \mathrm{z}_{1}=-1$

$\mathrm{AB}{2}=\left[\begin{array}{lll}2 & 0 & 1 \ 1 & 1 & 0 \ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{x}{2} \ \mathrm{y}{2} \ \mathrm{z}{2}\end{array}\right]=\left[\begin{array}{l}2 \ 3 \ 0\end{array}\right]$

$\mathrm{x}{2}=2, \mathrm{y}{2}=1, \mathrm{z}_{2}=-2$

$\mathrm{AB}{3}=\left[\begin{array}{lll}2 & 0 & 1 \ 1 & 1 & 0 \ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{x}{3} \ \mathrm{y}{3} \ \mathrm{z}{3}\end{array}\right]=\left[\begin{array}{l}3 \ 2 \ 1\end{array}\right]$

$\mathrm{x}{3}=2, \mathrm{y}{3}=0, \mathrm{z}_{3}=-1$

$\mathrm{B}=\left[\begin{array}{ccc}1 & 2 & 2 \ -1 & 1 & 0 \ -1 & -2 & -1\end{array}\right]$

$\alpha=|B|=3$

$\beta=1$

$\alpha^{3}+\beta^{3}=27+1=28$