if Ai= [ai bi and |a| 1 ,|b| 1 then show that sigma i=1 to infinity |Ai|= a2-b2/(1-a2)(1-b2) bi - Maths - Determinants

Question

if Ai= [ai bi and |a|<1 ,|b|
<1 then show that sigma i=1 to infinity |Ai|=
a2-b2/(1-a2)(1-b2)
bi ai ]

Akhil Goyal · Accepted Answer

Given:Ai=aibibiai⇒Ai=a2i-b2iTo prove:∑i=i∞Ai=a2-b2(1-a2)(1-b2)LHS=∑a2i-b2i=∑i=i∞a2i - ∑i=i∞b2iSince a, b<1, we can write the sum of these individual GPs=a21-a2-b21-b2=a2-b2(1-a2)(1-b2)=RHSHence Proved

if Ai= [ai bi and |a|<1 ,|b| <1 then show that sigma i=1 to infinity |Ai|= a2-b2/(1-a2)(1-b2) bi ai ]

if A_i= [aⁱ bⁱ and |a|<1 ,|b| <1 then show that sigma i=1 to infinity |A_i|= a²-b²/(1-a²)(1-b²)
bⁱ aⁱ ]