Are there any generalized inverses that would produce a left inverse for a short rectangular matrix? To give some context, I'm trying to solve the following problem:
$y = BA^{-1}x$
where:
$y$ = $n \times 1$ vector -- is known
$x$ = $3 \times 1$ vector -- is unknown
$B$ = $3 \times n$ matrix -- is known
$A$ = $n \times n$ singular matrix -- So $A$ is known and cannot be inverted to solve the problem.
If I could compute a generalized inverse for $B$ (let's call it $B^\dagger$) than I could avoid dealing with $A^{-1}$ altogether. The solution would simply be:
$x = AB^\dagger y$
However for this to work $B^\dagger$ would have to be a left inverse, i.e. $B^\dagger B = I$.
Any ideas on a generalized inverse method that would produce a left inverse for a short (less rows than columns) matrix? Is this even possible?
 A: Forgive my confusion. The question states $\mathbf{A}$ is singular (a classic inverse does not exist), yet the formula uses $\mathbf{A}^{-1}$ (the classic inverse). The answers are for different assumptions.
If the matrix product $\mathbf{C}=\mathbf{B}\,\mathbf{A}$, is known, build the psuedoinverse from the singular value decomposition:
$$
\mathbf{C} = \mathbf{U}\, \Sigma\, \mathbf{V}^{*} 
\qquad \Rightarrow \qquad
\mathbf{C}^{\dagger} = \mathbf{V}\, \Sigma^{\dagger} \,\mathbf{U}^{*}
$$
Both matrices $\mathbf{A}$ and $\mathbf{B}$ will also have a pseudoinverse. General comments follow.
Classification of inverses
Start $\mathbf{A}\in\mathbb{C}^{m\times m}_{m}$. The matrix is square and has full rank. Therefore the classic inverse $\mathbf{A}^{-1}$ exists and satisfies two equalities:
$$
\mathbf{A}^{-1}\mathbf{A} = \mathbf{A}\mathbf{A}^{-1} = \mathbf{I}_{m}
$$
The classic inverse is both a left and a right inverse
$$
\mathbf{A}^{-L}\mathbf{A} = \mathbf{A}\mathbf{A}^{-R} = \mathbf{I}_{m}
$$
What if we can only satisfy one of these inequalities? By relaxing the requirement, we generalize the notion of an inverse. The least squares problem naturally develops the Moore-Penrose pseudoinverse, $\mathbf{A}^{+}$.
If $\mathbf{A}^{+} \mathbf{A}= \mathbf{I}_{n}$, then the pseudoinverse $\mathbf{A}^{+}$ is a left inverse.  When $\mathbf{A} \mathbf{A}^{+}= \mathbf{I}_{n}$, the pseudoinverse $\mathbf{A}^{+}$ is a right inverse. When the pseuodinverse is a left and right inverse, the pseudoinverse in the classic inverse.
The remaining case is when the pseudoinverse is neither a left nor a right inverse. There is no identity inequality. The pseudoinverse is a projector on the column space of $\mathbf{A}$.

Watch the null spaces
Row and column rank deficient
For $\mathbf{A}\in\mathbb{C}^{m\times n}_{\rho}$, the singular value decomposition is
$$
\begin{align}
  \mathbf{A} &=
  \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}_{\rho\times \rho} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
\end{align}
$$
The Moore-Penrose pseudoinverse is
$$
\begin{align}
  \mathbf{A}^{+} &=
  \mathbf{V} \, \Sigma^{+} \, \mathbf{U}^{*} \\
%
 &=
% V 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}^{-1}_{\rho\times \rho} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% U* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
\end{align}
$$
When $\rho<\min\left(m,n\right)$, both null spaces are nontrivial.
The product matrices are
$$
\begin{align}
%
\mathbf{A}^{+} \mathbf{A} &=
% V 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{I}_{\rho} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  
%
=
%
\color{blue}{\mathbf{V}_{\mathcal{R}}}
\color{blue}{\mathbf{V}^{*}_{\mathcal{R}}}
%
\ne
%
\mathbf{I}_{n} \\
%%%%%
%
\mathbf{A} \mathbf{A}^{+} &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{I}_{\rho} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% U* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array} \right]  
%
=
%
\color{blue}{\mathbf{U}_{\mathcal{R}}}
\color{blue}{\mathbf{U}^{*}_{\mathcal{R}}}
%
\ne
%
\mathbf{I}_{m}
%
\end{align}
$$

_Row rank deficient, full column rank_
$$
  \mathbf{A}\in\mathbb{C}^{m\times n}_{n}, \quad
\mathbf{A} =
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{c}
     \mathbf{S}_{n\times n}  \\
     \mathbf{0}  
  \end{array} \right]
% V* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*}
  \end{array} \right], \quad
%%%
  \mathbf{A}^{+} =
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{c}
     \mathbf{S}^{-1}_{n\times n}  &
     \mathbf{0}  
  \end{array} \right]
% V* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*}
   & \color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array} \right]
$$
When $\rho=n<m$, only the right null space is trivial:
$$
\color{red}{\mathcal{N}\left( \mathbf{A}\right)} = \left\{ \mathbf{0} \right\}
$$
The product matrices are
$$
\begin{align}
%
\mathbf{A}^{+} \mathbf{A} &=
% V 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} 
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
    \mathbf{I}_{n}
  \end{array} \right]
% V* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} 
  \end{array} \right]  
%
=
%
\mathbf{I}_{n} \\
%%%%%
%
\mathbf{A} \, \mathbf{A}^{+} &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} &
     \color{red} {\mathbf{U}_{\mathcal{N}}} 
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{I}_{n} \\ \mathbf{0}
  \end{array} \right]  
% U* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\
     \color{red} {\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array} \right]  
%
=
%
\color{blue}{\mathbf{U}_{\mathcal{R}}}
\color{blue}{\mathbf{U}^{*}_{\mathcal{R}}}
%
\ne
%
\mathbf{I}_{m}
%
\end{align}
$$
This pseudoinverse is a left inverse: $\mathbf{A}^{+} = \mathbf{A}^{-L}$.

_Column rank deficient, full row rank_
$$
\mathbf{A}\in\mathbb{C}^{m\times n}_{m}, \quad
\begin{align}
  \mathbf{A} =
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} 
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}_{m\times m} & \mathbf{0} 
  \end{array} \right]
% V* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  , \quad
  \mathbf{A}^{+} =
% V 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} 
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}^{-1}_{m\times m}  \\
     \mathbf{0} 
  \end{array} \right]
% U* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
\end{align}
$$
When $\rho=m<n$, only the left null space is trivial,
$$
\color{red}{\mathcal{N}\left( \mathbf{A}^{*}\right)} = \left\{ \mathbf{0} \right\}
$$
The product matrices are
$$
\begin{align}
%
\mathbf{A}^{+} \mathbf{A} &=
% V 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} 
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{I}_{m} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} 
  \end{array} \right]  
=
\color{blue}{\mathbf{V}_{\mathcal{R}}}
\color{blue}{\mathbf{V}^{*}_{\mathcal{R}}}
%
\ne
%
\mathbf{I}_{n} \\
%%%%%
%
\mathbf{A}\, \mathbf{A}^{+} &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} 
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
    \mathbf{I}_{m}
  \end{array} \right]  
% U* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*}
  \end{array} \right]  
%
=
\color{blue}{\mathbf{U}_{\mathcal{R}}}
\color{blue}{\mathbf{U}^{*}_{\mathcal{R}}}
%
= \mathbf{I}_{m}
%
\end{align}
$$
Therefore, the pseudoinverse is a right inverse: $\mathbf{A}^{+} = \mathbf{A}^{-R}$.

_Full column rank deficient, full row rank_
$$
\mathbf{A}\in\mathbb{C}^{m\times m}_{m}, \quad
\begin{align}
  \mathbf{A} =
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} 
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}_{m\times m} 
  \end{array} \right]
% V* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*}
  \end{array} \right]  , \quad
  \mathbf{A}^{+} =
% V 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} 
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}^{-1}_{m\times m}
  \end{array} \right]
% U* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} 
  \end{array} \right]  \\
%
\end{align}
$$
When $\rho=m$, both null spaces are trivial,
$$
\begin{align}
%
\color{red}{\mathcal{N}\left( \mathbf{A}\right)} &= \left\{ \mathbf{0} \right\} \\
%
\color{red}{\mathcal{N}\left( \mathbf{A}^{*}\right)} &= \left\{ \mathbf{0} \right\}
\end{align}
%
$$
The product matrices are
$$
\begin{align}
%
\mathbf{A}^{+} \mathbf{A} &= \mathbf{I}_{m} \\
%%%%%
%
\mathbf{A}\, \mathbf{A}^{+} &= \mathbf{I}_{m}
%
\end{align}
$$
Therefore, the pseudoinverse is both a right inverse: $\mathbf{A}^{+} = \mathbf{A}^{-R}$ and a left inverse: $\mathbf{A}^{+} = \mathbf{A}^{-L}$.
