There is a subject which I’ve mentioned in scant ways, in other postings, but which I have not yet written a posting about, directly.

A situation can arise in which we have a rotation matrix, and where we’d like to invert the rotation it defines, but where the task of actually computing the inverse of the matrix would be too expensive computationally. And so the question can come up, of whether to use the transpose will also work.

The answer is, that in one specific situation the transpose can be used. This is, if each column of the matrix is a unit vector, and if all the columns are mutually orthogonal. If that’s the case, then the determinant of the matrix will also be equal to (1), and the transpose will be equal to the inverse.

Simultaneously, this will then also be true for all *the rows* of (either) matrix.

Therefore, when I wrote that a Perpendicular matrix could be computed, in case the original matrix defined a quadric, and that the transpose of that perpendicular matrix can be used, as if that was its inverse, I was describing a special situation, in which the perpendicular matrix would have been orthonormal to begin with.

In case a matrix which defines a conversion of coordinates is not orthonormal, then this approach will not work, and its inverse must be used. And I think that the resulting matrix is also called ‘The Jordan Product’ of an original matrix. My exercise did not need this.

Dirk