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Is there a "simple" formula for computing the Moore-Penrose pseudoinverse of a $3\times 3$ matrix? I mean something like the formula for the inverse (for non-singular matrices), which involves the matrix of minors, etc.
I need that for a computer program, and I feel that using LAPACK's SVD is a bit of an overkill.
You are probably thinking of the formula $$ \begin{align} \mathbf{A}^{-1} &= \frac{\text{adj } \mathbf{A}} {\det \mathbf{A} } \\ &= \frac{\left( \text{cof } \mathbf{A}\right)^{\mathrm{T}}} {\det \mathbf{A} } \\ \end{align} $$ The matrix $\text{adj } \mathbf{A}$ is the adjugate of $\mathbf{A}$ and is the transpose of $\mathbf{C}$, the matrix of cofactors of $\mathbf{A}$.
For a nonsingular $\mathbf{A}\in\mathbb{R}^{3 x 3}$, $$ \mathbf{A} = \left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right], $$ the matrix of cofactors is composed of the determinants $$ \mathbf{C} = \left[ \begin{array}{ccc} % + \left| \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} \right| & - \left| \begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} \right| & + \left| \begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} \right| \\ % - \left| \begin{array}{cc} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array} \right| & + \left| \begin{array}{cc} a_{11} & a_{13} \\ a_{31} & a_{33} \end{array} \right| & - \left| \begin{array}{cc} a_{11} & a_{12} \\ a_{31} & a_{32} \end{array} \right| \\ % + \left| \begin{array}{cc} a_{12} & a_{13} \\ a_{22} & a_{23} \end{array} \right| & - \left| \begin{array}{cc} a_{11} & a_{13} \\ a_{21} & a_{23} \end{array} \right| & + \left| \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| \\ \end{array} \right]. $$
Because you specified that $\mathbf{A}$ is nonsingular, the matrix inverse exists and is the same as the Moore-Penrose pseudoinverse: $$ \mathbf{A}^{-1} = \mathbf{A}^{\dagger}. $$
Example Pencil and paper exercise of confirmation: $$ \mathbf{A} = \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{array} \right], $$ The determinant is $\det \mathbf{A} = 1$, and the matrix of cofactors is $$ \mathbf{C} = \left[ \begin{array}{rrr} 0 & 1 & -1 \\ 0 & -1 & 0 \\ -1 & -1 & 1 \end{array} \right] $$ The inverse matrix is $$ \mathbf{A}^{-1} = \frac{\left( \text{cof } \mathbf{A}\right)^{\mathrm{T}}} {\det \mathbf{A} } = \frac{\mathbf{C}^\mathrm{T}} {-1} = \left[ \begin{array}{rrr} 0 & 0 & 1 \\ -1 & \phantom{-}1 & 1 \\ 1 & 0 & -1 \end{array} \right]. $$
