PX153 - K8 - properties of matrix inverses

$(i) :$ if $A$ and $B$ are invertible (have inverses), and have the shape, $n \times n$ , then $A B$ is also invertible, and $(A B)^{- 1} = B^{- 1} A^{- 1}$
- proof: $(B^{- 1} A^{- 1}) (A B) = B^{- 1} (A^{- 1} A) B = B^{- 1} (I) B = I$
$(i i) :$ if $A$ is a $n \times n$ invertible matrix, then $A^{T}$ is invertible, and $(A^{T})^{- 1} = (A^{- 1})^{T}$
- proof:

\begin{aligned} ((A^{- 1})^{T} A^{T})_{i k} & = \sum_{j = 1}^{n} (a^{- 1})_{i j}^{T} (a)_{j k}^{T} \\ = \sum_{j = 1}^{n} (a^{- 1})_{j i} (a)_{k j} \\ = \sum_{j = 1}^{n} a_{k j} a_{j i}^{- 1} \\ = (A A^{- 1})_{k i} \\ = (I)_{k i} \\ = (I)_{i k} \end{aligned}

in principle, the inverse of a matrix can be used to compute solutions to simultaneous equations
- eg: to solve $A \vec{x} = \vec{b} :$ if $A^{- 1}$ exists, then $\vec{x} = A^{- 1} \vec{b}$
earlier, gaussian elimination was used from interchanges of lines (rows)
the inverse method of finding $A^{- 1}$ and solving for $\vec{x}$ needs of the order of $O (n \times n!)$ , which is very inefficient for $n > 3$
methods using gaussian elimination are actually faster in most cases