Welcome to mathimatikoi.org forum; Enjoy your visit here.

## An equality with matrices

Mathematical Competitions
Riemann
Articles: 0
Posts: 174
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

### An equality with matrices

Let $A, B$ be $3 \times 3$ matrices with real entries. Prove that

$$A - \left ( A^{-1} +\left ( B^{-1} - A \right )^{-1} \right )^{-1} = ABA$$

provided all the inverses appearing on the left hand side exist.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Tolaso J Kos
Articles: 2
Posts: 861
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

### Re: An equality with matrices

Let $A, B$ be elements of an arbitrary associative algebra with unit. Then:

\begin{align*}
\left ( A^{-1} +\left ( B^{-1} - A \right )^{-1} \right )^{-1} &= \left ( A^{-1} \left ( B^{-1} - A \right )\left ( B^{-1} - A \right )^{-1} + A^{-1} A \left ( B^{-1} - A \right )^{-1} \right )^{-1} \\
&=\left ( A^{-1} \left ( \left ( B^{-1} - A \right ) +A \right )\left ( B^{-1} -A \right )^{-1} \right )^{-1} \\
&= \left ( A^{-1} B^{-1} \left ( B^{-1} -A \right )^{-1} \right )^{-1}\\
&= \left ( B^{-1}-A \right ) BA \\
&= A - ABA
\end{align*}
Imagination is much more important than knowledge.