उत्तर (28)

समाधान

$A=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right] \quad B=\left[B _1, B _2, B _3\right]$

$B _1=\left[\begin{array}{l}x _1 \\ y _1 \\ z _1\end{array}\right], \quad B _2=\left[\begin{array}{l}x _2 \\ y _2 \\ z _2\end{array}\right], \quad B _3=\left[\begin{array}{l}x _3 \\ y _3 \\ z _3\end{array}\right]$

$AB _1=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x _1 \\ y _1 \\ z _1\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$

$x _1=1, y _1=-1, z _1=-1$

$AB _2=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x _2 \\ y _2 \\ z _2\end{array}\right]=\left[\begin{array}{l}2 \\ 3 \\ 0\end{array}\right]$

$x _2=2, y _2=1, z _2=-2$

$AB _3=\left[\begin{array}{lll}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x _3 \\ y _3 \\ z _3\end{array}\right]=\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]$

$x _3=2, y _3=0, z _3=-1$

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

$\alpha=|B|=3$

$\beta=1$

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