From B=(I+A)−1 we get B(I+A)=I, so BA=I−B.
From A+C=I: C=I−A.
Calculating CB=C(I+A)−1=(I−A)(I+A)−1.
Since B(I+A)=I, we have (I+A)=B−1, so C=(I−A)B−1.
Multiplying by B: CB=(I−A).
However, using the given BC=[1−1−52] and B(I+A)=I implies B−BA=B−(I−B)=2B−I=BC, i.e. BC=2B−I.
Therefore B=21(BC+I)=21[2−1−53]=[1−1/2−5/23/2].
Also CB=2B−I=[1−1−52].
From CB[x1x2]=[12−6]: (CB)−1=−31[2151].
So [x1x2]=−31[−66]=[2−2].
Therefore x1+x2=0