Let satisfy for some . Then is equal to _______. [2026]

Question

Let $A = [\begin{matrix} - 1 & 1 & - 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}]$ satisfy $A^{2} + α (adj (adj (A))) + β (adj (A) (adj (adj (A)))) = [\begin{matrix} 2 & - 2 & 2 \\ - 2 & 0 & - 1 \\ 0 & 0 & - 1 \end{matrix}]$ for some $α, β \in R$ . Then ${(α - β)}^{2}$ is equal to _______. [2026]

Smart Achievers · Accepted Answer

(4)

$A = [\begin{matrix} - 1 & 1 & - 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}]$

$\Rightarrow | A | = - 1$

$∵ adj (adj A) = | A |^{n - 2} A$

and $A a d j A = | A | I$

Now, $A^{2} = [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

We have, $A^{2} + α (adj (adj (A)) + β (adj (A) (adj (adj (A))) = [\begin{matrix} 2 & - 2 & 2 \\ - 2 & 0 & - 1 \\ 0 & 0 & - 1 \end{matrix}]$

$\Rightarrow A^{2} + α (- A) + β I = [\begin{matrix} 2 & - 2 & 2 \\ - 2 & 0 & - 1 \\ 0 & 0 & - 1 \end{matrix}]$

$\Rightarrow [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 0 & - 1 \\ 0 & 0 & - 1 \end{matrix}] - α [\begin{matrix} - 1 & 1 & - 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}] + β = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & - 2 & 2 \\ - 2 & 0 & - 1 \\ 0 & 0 & - 1 \end{matrix}]$

$\Rightarrow - [\begin{matrix} 2 + α + β & - 1 - α & 1 + α \\ - 1 - α & 1 + β & - α \\ 0 & 0 & 1 - α + β \end{matrix}] = [\begin{matrix} 2 & - 2 & 2 \\ - 2 & 0 & - 1 \\ 0 & 0 & - 1 \end{matrix}]$

By comparison on both sides, we get

$α = 1 and 1 + β = 0 \Rightarrow β = - 1$

$∴ {(α - β)}^{2} = {(1 + 1)}^{2} = 2^{2} = 4$

Solution

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Solution

Q. Let A=[–11–1101001] satisfy A2+α(adj(adj(A)))+β(adj(A)(adj(adj(A))))=[2–22–20–100–1] for some α,β∈R. Then (α–β)2 is equal to _______. [2026]

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