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 _____________. - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Solution:

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

Solution:

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