Let A=1-12α and B= β110,α,β∈R. Let α1 be the value of α which satisfies A+B2=A2+2222 and α2 be the value of α which satisfies A+B2=B2. Then α1-α2 is equal to _____________.

A+B=β+103α ⇒A+B2=β+103αβ+103α =β+1203β+1+3αα2, A2=1-12α1-12α =-1-1-α2+2αα2-2 ∴1-α+12α+4α2=β+1203α+β+1α2 α=1⇒α1=1, B2=β110β110 =β2+1ββ1=β+1203β+1+3αα2 ∴β=0, α=-1=α2, α1-α2=1--1=2

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$L e t A = [\begin{matrix} 1 & - 1 \\ 2 & α \end{matrix}] a n d B = [\begin{matrix} β & 1 \\ 1 & 0 \end{matrix}], α, β \in R . L e t α_{1} b e t h e v a l u e o f α w h i c h s a t i s f i e s {(A + B)}^{2} = A^{2} + [\begin{matrix} 2 & 2 \\ 2 & 2 \end{matrix}] a n d α_{2} b e t h e v a l u e o f α w h i c h s a t i s f i e s {(A + B)}^{2} = B^{2} . T h e n |α_{1} - α_{2}| i s e q u a l t o_____________.$

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Detailed Solution

$A + B = [\begin{matrix} β + 1 & 0 \\ 3 & α \end{matrix}] \Rightarrow {(A + B)}^{2} = [\begin{matrix} β + 1 & 0 \\ 3 & α \end{matrix}] [\begin{matrix} β + 1 & 0 \\ 3 & α \end{matrix}] = [\begin{matrix} {(β + 1)}^{2} & 0 \\ 3 (β + 1) + 3 α & α^{2} \end{matrix}], A^{2} = [\begin{matrix} 1 & - 1 \\ 2 & α \end{matrix}] [\begin{matrix} 1 & - 1 \\ 2 & α \end{matrix}] = [\begin{matrix} - 1 & - 1 - α \\ 2 + 2 α & α^{2} - 2 \end{matrix}] ∴ [\begin{matrix} 1 & - α + 1 \\ 2 α + 4 & α^{2} \end{matrix}] = [\begin{matrix} {(β + 1)}^{2} & 0 \\ 3 (α + β + 1) & α^{2} \end{matrix}] \underset{}{α = 1 \Rightarrow α_{1} = 1,} B^{2} = [\begin{matrix} β & 1 \\ 1 & 0 \end{matrix}] [\begin{matrix} β & 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} β^{2} + 1 & β \\ β & 1 \end{matrix}] = [\begin{matrix} {(β + 1)}^{2} & 0 \\ 3 (β + 1) + 3 α & α^{2} \end{matrix}] ∴ β = 0, α = - 1 = α_{2}, |α_{1} - α_{2}| = |1 - (- 1)| = 2$