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I've generated a 3 by 5 matrix using Matlab:

$$ \begin{pmatrix} 21 & 3 & 4 & 4 & 17 \\ 23 & 7 & 25 & 11 & 1 \\ 4 & 14 & 24 & 23 & 22 \\ \end{pmatrix} $$

A right inverse to this matrix is

$$ \begin{pmatrix} 0.0315 & 0.0191 & -0.0252\\ 0 & 0 & 0\\ -0.0300 & 0.0236 & 0.0221\\ 0 & 0 & 0\\ 0.0270 & -0.0292 & 0.0259\\ \end{pmatrix} $$

This inverse was found using this method: column 1 of the inverse is equal to the solution of $$ Ax=\begin{pmatrix}1\\0\\0\\\end{pmatrix}$$ Column 2 of the inverse is the solution to $$ Ax=\begin{pmatrix}0\\1\\0\\\end{pmatrix}$$ And so on.

My question is the following: why does the inverse have two empty rows? The method of generating the inverse has consistently yielded a number of empty rows equal to the number of columns in the original matrix minus its rank. The empty rows are not always at the same row as in this example (i.e. they are not always on rows 2 and four).

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Hint : The entries of the second and fourth column are irrelevant. Calculate the inverse of the matrix $$\pmatrix{21&4&17\\23&25&1\\4&24&22}$$ You get the calculated matrix without the zeros.

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    Right: it seems that Matlab just chose three random columns to act as the pivots. I took the inverse of the first three columns and added two empty rows: this worked as another right inverse. – 2017-02-05
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It seems that it is essentially the method used by mathlab. Your generated matrix is of rank 3 then it can accept a right inverse with two zero (or empty) rows.