Let A=[ … ], where B_1, B_2,B_3 are column matrices, and AB_1=[ … ], AB_2=[ … ] If α =|B| and β is the sum of all the diagonal elements of B, then α…

Given: $A=[\begin{matrix}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{matrix}]$A=[211​010​101​], $B=[\begin{matrix}{B}_{1} & {B}_{2} & {B}_{3}\end{matrix}]$B=[B1​​B2​​B3​​], ${B}_{1}=[\begin{matrix}{x}_{1} \\ {y}_{1} \\ {z}_{1}\end{matrix}]$B1​=[x1​y1​z1​​], ${B}_{2}=[\begin{matrix}{x}_{2} \\ {y}_{2} \\ {z}_{2}\end{matrix}]$B2​=[x2​y2​z2​​] and ${B}_{3}=[\begin{matrix}{x}_{3} \\ {y}_{3} \\ {z}_{3}\end{matrix}]$B3​=[x3​y3​z3​​].

$$\Rightarrow {\mathrm{AB}}_{1}=[\begin{matrix}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{matrix}][\begin{matrix}{x}_{1} \\ {y}_{1} \\ {z}_{1}\end{matrix}]=[\begin{matrix}1 \\ 0 \\ 0\end{matrix}]$$⇒AB1​=[211​010​101​][x1​y1​z1​​]=[100​]

$$\Rightarrow [\begin{matrix}2{x}_{1}+0+{z}_{1} \\ {x}_{1}+{y}_{1}+0 \\ {x}_{1}+0+{z}_{1}\end{matrix}]=[\begin{matrix}1 \\ 0 \\ 0\end{matrix}]$$⇒[2x1​+0+z1​x1​+y1​+0x1​+0+z1​​]=[100​]

$$\Rightarrow 2{x}_{1}+{z}_{1}=1,{x}_{1}+{y}_{1}=0,{x}_{1}+{z}_{1}=0$$⇒2x1​+z1​=1,x1​+y1​=0,x1​+z1​=0

$$\Rightarrow {x}_{1}=1,{y}_{1}=-1,{z}_{1}=-1$$⇒x1​=1,y1​=−1,z1​=−1

$$\Rightarrow {\mathrm{AB}}_{2}=[\begin{matrix}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{matrix}][\begin{matrix}{x}_{2} \\ {y}_{2} \\ {z}_{2}\end{matrix}]=[\begin{matrix}2 \\ 3 \\ 0\end{matrix}]$$⇒AB2​=[211​010​101​][x2​y2​z2​​]=[230​]

$$\Rightarrow {\mathrm{AB}}_{2}=[\begin{matrix}2{x}_{2}+{z}_{2} \\ {x}_{2}+{y}_{2} \\ {x}_{2}+{z}_{2}\end{matrix}]=[\begin{matrix}2 \\ 3 \\ 0\end{matrix}]$$⇒AB2​=[2x2​+z2​x2​+y2​x2​+z2​​]=[230​]

$$\Rightarrow 2{x}_{2}+{z}_{2}=2,{x}_{2}+{y}_{2}=3,{x}_{2}+{z}_{2}=0$$⇒2x2​+z2​=2,x2​+y2​=3,x2​+z2​=0

$$\Rightarrow {x}_{2}=2,{y}_{2}=1,{z}_{2}=-2$$⇒x2​=2,y2​=1,z2​=−2

$$\Rightarrow {\mathrm{AB}}_{3}=[\begin{matrix}2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{matrix}][\begin{matrix}{x}_{3} \\ {y}_{3} \\ {z}_{3}\end{matrix}]=[\begin{matrix}3 \\ 2 \\ 1\end{matrix}]$$⇒AB3​=[211​010​101​][x3​y3​z3​​]=[321​]

$$\Rightarrow {\mathrm{AB}}_{2}=[\begin{matrix}2{x}_{3}+{z}_{3} \\ {x}_{3}+{y}_{3} \\ {x}_{3}+{z}_{3}\end{matrix}]=[\begin{matrix}3 \\ 2 \\ 1\end{matrix}]$$⇒AB2​=[2x3​+z3​x3​+y3​x3​+z3​​]=[321​]

$$\Rightarrow 2{x}_{3}+{z}_{3}=3,{x}_{3}+{y}_{3}=2,{x}_{3}+{z}_{3}=1$$⇒2x3​+z3​=3,x3​+y3​=2,x3​+z3​=1

$$\Rightarrow {x}_{3}=2,{y}_{3}=0,{z}_{3}=-1$$⇒x3​=2,y3​=0,z3​=−1

$$\Rightarrow B=[\begin{matrix}1 & 2 & 2 \\ -1 & 1 & 0 \\ -1 & -2 & -1\end{matrix}]$$⇒B=[1−1−1​21−2​20−1​]

$$\Rightarrow \alpha =|B|=3$$⇒α=∣B∣=3

$$\Rightarrow \beta =1$$⇒β=1

$$\Rightarrow {\alpha }^{3}+{\beta }^{3}=27+1=28$$⇒α3+β3=27+1=28