Let X=(010001000),Y=αI+βX+γX2 and Z=α2I−αβX+(β2−αγ)X2 ,α,β,γ∈R . If Y−1=(15−2515015−250015) then (α+β+γ)2 is equal to _____.

X=(010001000),X2=(001000000) Y=(αβγ0αβ00α),Z=(α2−αββ2−αγ0α2−αβ00α2) Y.Y−1=I (αβγ0αβ00α)(15−2515015−250015)=(010001000) α5=1⇒α=5 −25α+β5=0⇒β=10 α5−2β5+γ5=0 ⇒γ=15 ⇒(α+β+γ)2=(5+10+15)2=900

Let X=(010001000),Y=αI+βX+γX2 and Z=α2I−αβX+(β2−αγ)X2 ,α,β,γ∈R . If Y−1=(15−2515015−250015) then (α+β+γ)2 is equal to _____.

X=(010001000),X2=(001000000) Y=(αβγ0αβ00α),Z=(α2−αββ2−αγ0α2−αβ00α2) Y.Y−1=I (αβγ0αβ00α)(15−2515015−250015)=(010001000) α5=1⇒α=5 −25α+β5=0⇒β=10 α5−2β5+γ5=0 ⇒γ=15 ⇒(α+β+γ)2=(5+10+15)2=900

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Let $X = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}), Y = α I + β X + γ X^{2}$ and $Z = α^{2} I - α β X + (β^{2} - α γ) X^{2}, α, β, γ \in R$ . If $Y^{- 1} = (\begin{matrix} \frac{1}{5} & \frac{- 2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & \frac{- 2}{5} \\ 0 & 0 & \frac{1}{5} \end{matrix})$ then ${(α + β + γ)}^{2}$ is equal to _____.

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Detailed Solution

$X = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}), X^{2} = (\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) Y = (\begin{matrix} α & β & γ \\ 0 & α & β \\ 0 & 0 & α \end{matrix}), Z = (\begin{matrix} α^{2} & - α β & β^{2} - α γ \\ 0 & α^{2} & - α β \\ 0 & 0 & α^{2} \end{matrix}) Y . Y^{- 1} = I (\begin{matrix} α & β & γ \\ 0 & α & β \\ 0 & 0 & α \end{matrix}) (\begin{matrix} \frac{1}{5} & \frac{- 2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & \frac{- 2}{5} \\ 0 & 0 & \frac{1}{5} \end{matrix}) = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}) \frac{α}{5} = 1 \Rightarrow α = 5 - \frac{2}{5} α + \frac{β}{5} = 0 \Rightarrow β = 10 \frac{α}{5} - \frac{2 β}{5} + \frac{γ}{5} = 0 \Rightarrow γ = 15 \Rightarrow {(α + β + γ)}^{2} = {(5 + 10 + 15)}^{2} = 900$