Given A=[12−1α] and B=[β110],α,β∈R,
So, A+B=[β+130α]
Now (A+B)2=[β+130α][β+130α]
=[(β+1)23(β+1)+3α0α2]
Also, A2=[12−1α][12−1α]=[−12+2α−1−αα2−2]
Now solving (A+B)2=A2+[2222]
⇒[(β+1)23(α+β+1)0α2]=[12α+4−α+1α2]
Now on comparing both side we get, α=1=α1
And B2=[β110][β110]=[β2+1ββ1]
Now using (A+B)2=B2
⇒[β2+1ββ1]=[(β+1)23(β+1)+3α0α2]
Again on comparing both side we get, β=0,α=−1=α2
So, ∣α1−α2∣=∣1−(−1)∣=2