(A) 4
(B) 2
(C) 0
(D) −2
Correct Answer: 0
B = (I + A)−1 and A + C = I
From A + C = I, we get
C = I − A
Now,
CB = (I − A)B = B − AB
Since,
(I + A)B = I
⇒ AB = I − B
Substitute:
CB = B − (I − B) = 2B − I
Thus CB is fully known.
Solving CB [ x1 x2 ] = [ 12 −6 ] gives
x1 + x2 = 0