Matrix equality with identity 
Given two matrices $A$ and $B$. To show:
Part 1: $(I+AB)^{-1}=I-A(I+BA)^{-1}B$, provided $BA$ and $AB$ exist.
Part 2: Let $a$ and $b$ be scalars. Using part 1, to show $(aI_{k}+bJ_{k})^{-1}=I_{k}/a -bJ_{k}/\{a(a+kb)\}$, where $I_p$ is the $p\times p$ identity matrix, and $J_p$ is the $p\times p$-matrix whose all entries are $1$.

My approach:
1st part:
$(I-A(I+BA)^{-1}B)(I+AB)=(I+AB)-A(I+BA)^{-1}B(I+AB)$
Since, $B(I+AB)=B+BAB=(I+BA)B$
$(I-A(I+BA)^{-1}B)(I+AB)=(I+AB)-A(I+BA)^{-1}(I+BA)B=I$
Similarly, $(I+AB)(I-A(I+BA)^{-1}B)=I$
2nd part:
We have $aI_{k}+bJ_{k}=a(I_k+\frac{b}{a}1_k1_k^{\prime})$, where $1_k$ is the $k\times 1$ vector filled with ones.
Fix, $\frac{b}{a}1_k=A, 1_k^{\prime}=B$
We get,
$(aI_{k}+bJ_{k})^{-1}=\frac{1}{a}(I_k+\frac{b}{a}1_k1_k^{\prime})^{-1}$
$=\frac{1}{a}(I_k-\frac{b}{a}1_{k}(1+\frac{kb}{a})^{-1}1_k^{\prime})$
$=\frac{1}{a}(I_k-\frac{ab}{a(a+kb)}J_k)$
$=\frac{1}{a}(I_k-\frac{b}{a(a+kb)}J_k)$
I was wondering if you could review my work or suggest any alternative way to prove these problems. I appreciate your feedback!

You write $B(I+AB)=B+BAB=(I+AB)B$, but the last term should be $(I+BA)B$ (the order of $A$ and $B$ matters since $AB \neq BA$ in general).
I can't comment on the second part because I'm not sure what is meant by $J_p$ and $1_p$. 
