Let A = [ 2 0 1 1 1 0 1 0 1 ] , B = [ B 1 , B 2 , B 3 ] , where B 1 , B 2 , B 3 are column matrices, and A B 1 = [ 1 0 0 ] , A B 2 = [ 2 3 0 ] , A B 3 = [ 3 2 1 ] . If = | B | and is the sum of all the diagonal elements of B, then 3 +3 is e

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Solution

Q.
Let $A = [\begin{matrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}],$ $B = [\begin{matrix} B_{1}, & B_{2}, & B_{3} \end{matrix}]$ , where $B_{1}, B_{2}, B_{3}$ are column matrices, and $A B_{1} = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}], A B_{2} = [\begin{matrix} 2 \\ 3 \\ 0 \end{matrix}], A B_{3} = [\begin{matrix} 3 \\ 2 \\ 1 \end{matrix}] .$ If $α = | B |$ and $β$ is the sum of all the diagonal elements of B, then $α^{3} + β^{3}$ is equal to _____ . [2024]

Ans.
(28)

Let $B_{1} = [\begin{matrix} a_{1} \\ b_{1} \\ c_{1} \end{matrix}],$ $A B_{1} = [\begin{matrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}] [\begin{matrix} a_{1} \\ b_{1} \\ c_{1} \end{matrix}] = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}]$

$\Rightarrow 2 a_{1} + c_{1} = 1, a_{1} + b_{1} = 0, a_{1} + c_{1} = 0$

On solving, we get $a_{1} = 1, b_{1} = - 1, c_{1} = - 1$ ...(i)

Similarly, let $B_{2} = [\begin{matrix} a_{2} \\ b_{2} \\ c_{2} \end{matrix}], A B_{2} = [\begin{matrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}] [\begin{matrix} a_{2} \\ b_{2} \\ c_{2} \end{matrix}] = [\begin{matrix} 2 \\ 3 \\ 0 \end{matrix}]$

$\Rightarrow 2 a_{2} + c_{2} = 2, a_{2} + b_{2} = 3$ and $a_{2} + c_{2} = 0$

On solving, we get $a_{2} = 2, b_{2} = 1$ and $c_{2} = - 2$ ...(ii)

Similarly, let $B_{3} = [\begin{matrix} a_{3} \\ b_{3} \\ c_{3} \end{matrix}], A B_{3} = [\begin{matrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}] [\begin{matrix} a_{3} \\ b_{3} \\ c_{3} \end{matrix}] = [\begin{matrix} 3 \\ 2 \\ 1 \end{matrix}]$

$\Rightarrow 2 a_{3} + c_{3} = 3, a_{3} + b_{3} = 2$ and $a_{3} + c_{3} = 1$

$\Rightarrow a_{3} = 2, b_{3} = 0, c_{3} = - 1$ ...(iii)

Thus, $B = [\begin{matrix} 1 & 2 & 2 \\ - 1 & 1 & 0 \\ - 1 & - 2 & - 1 \end{matrix}]$ [By using (i), (ii) & (iii)]

$α = | B | = 1 (- 1 + 0) - 2 (1 - 0) + 2 (2 + 1) = - 1 - 2 + 6 = 3$

and $β =$ sum of diagonal elements of $B = 1 + 1 + (- 1) = 1$

Hence, $α^{3} + β^{3} = {(3)}^{3} + {(1)}^{3} = 28$

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