Given: A = \left[\begin{cases}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{cases}\right], B = \left[\begin{cases}{B}_{1} & {B}_{2} & {B}_{3}\end{cases}\right], {B}_{1} = \left[x1y1z1\right], {B}_{2} = \left[x2y2z2\right] and {B}_{3} = \left[x3y3z3\right].
⇒ {AB}_{1} = \left[\begin{cases}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{cases}\right]\left[x1y1z1\right] = \left[100\right]
⇒ \left[2x1+0+z1x1+y1+0x1+0+z1\right] = \left[100\right]
⇒2x1+z1=1,x1+y1=0,x1+z1=0
⇒x1=1,y1=−1,z1=−1
⇒ {AB}_{2} = \left[\begin{cases}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{cases}\right]\left[x2y2z2\right] = \left[230\right]
⇒ {AB}_{2} = \left[2x2+z2x2+y2x2+z2\right] = \left[230\right]
⇒2x2+z2=2,x2+y2=3,x2+z2=0
⇒x2=2,y2=1,z2=−2
⇒ {AB}_{3} = \left[\begin{cases}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{cases}\right]\left[x3y3z3\right] = \left[321\right]
⇒ {AB}_{2} = \left[2x3+z3x3+y3x3+z3\right] = \left[321\right]
⇒2x3+z3=3,x3+y3=2,x3+z3=1
⇒x3=2,y3=0,z3=−1
⇒ B = \left[\begin{cases}1 & 2 & 2 \\ - 1 & 1 & 0 \\ - 1 & - 2 & - 1\end{cases}\right]
⇒α=∣B∣=3
⇒β=1
⇒α3+β3=27+1=28
