Inverse not adding up 
I'm trying to get this to add up but in my 2nd multiplication I get:
$$-I(I-BC)^{-1} B + B(I+C(I-BC)^{-1} B)$$ $$\Rightarrow -I(XB)+B(I+CXB)$$ if I let $$(I-BC)^{-1} = X$$
But this does not add up to become 0 like it is supposed to?
In the last calculation I get:
$$C(-(I-BC)^{-1} B) + I(I+C(I-BC)^{-1} B)$$
$$\Rightarrow C(-XB)+I(I+CXB)$$
Now this will not add up to 1 as it is supposed to.
EDIT:
Directly after this exercise I am to do:

And I guess I am to use the formula for 66...The same one that I am doing...It seems to be the only one that adds up. Would that be correct? 
 A: $$
−I(I−BC)^{−1}B+B(I+C(I−BC)^{−1}B)=B+BC(I-BC)^{-1}B-(I-BC)^{-1}B\\
=B+(BC-I)(I-BC)^{-1}B=B-B=0.
$$
$$
C(−(I−BC)^{−1}B)+I(I+C(I−BC)^{−1}B)=-C(I-BC)^{-1}B+I+C(I-BC)^{-1}B=I.
$$
For 69, $B=0$ and so $BC=CB=0$; the inverse is 
$$
\begin{bmatrix} I&0\\-C&I\end{bmatrix}=\begin{bmatrix}1&0&0&0\\0&1&0&0\\-2&-3&1&0\\-1&-2&0&1\end{bmatrix}.
$$
A: Let's start at: $−I(XB)+B(I+CXB)$; we can multiply out to give $-IXB+B+BCXB$; matrix addition is commutative, so this is $-IXB+BCXB+B=-(I-BC)XB+B=-B+B=0$, because $X$ is the inverse of exactly the thing that it is being multiplied by in the parentheses.
(The same thing will happen with the fourth case: you've got a $CX$ and a $-CX$ (with an $I$ in between) and they will still cancel.)
A: I would just say
$−I(I−BC)^{−1}B+B(I+C(I−BC)^{−1}B)=$
$=−I(I−BC)^{−1}B+(B+BC(I−BC)^{−1}B)=$
$=(-(I−BC)^{−1} + BC(I−BC)^{−1})B +B$ by grouping now $(I−BC)^{−1} $
$=((I−BC)^{−1}(-I+BC))B+B =$
$-B +B= 0$
A: For problem 69 you can invent your own formula on the fly. Let $A$ denote the given matrix and partition $A$ as 
$A = \begin{pmatrix}
 I & 0 \\
 C & I
\end{pmatrix}$.
Now let 
$D=
\begin{pmatrix}
D_1 & D_2
\\ D_3 & D_4
\end{pmatrix}
$
be the the inverse, so that $AD=I$
or equivalently 
$$
\begin{pmatrix}
 I & 0 \\
 C & I
\end{pmatrix}
\begin{pmatrix}
D_1 & D_2
\\ D_3 & D_4
\end{pmatrix}
=
\begin{pmatrix}
I & 0
\\ 0 & I
\end{pmatrix}.
$$
Expanding the left hand side and equating submatrices we get 4 equations for what the entries of the inverse must look like:
$D_1= I $, $D_2=0$, $CD_1 + D_3 = 0$ and $CD_2 + D_4 = I.$ Plugging in for $D_1$ and $D_2$ we solve for the remaining matrices: $D_3 = - C$ and $D_4 = I.$ Thus the inverse $D$ must take the form 
$
D=
\begin{pmatrix}
 I & 0
\\ -C & I\end{pmatrix}$, which agrees structurally with the fact that the inverse of a unit lower triangular matrix is also unit lower triangular.
