Exercise 2: Why R^-1=R^T Holds for Rotation but Not for Scaling — Possible Solution
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WHY IT HOLDS SPECIFICALLY FOR ROTATION MATRICES
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This chapter's own explanation traces the R^-1=R^T shortcut back to
Chapter 4's dot-product orthogonality checks: a rotation matrix's
rows (and columns) are mutually perpendicular UNIT vectors - each row
has length exactly 1, and any two different rows have a dot product
of exactly 0. This specific structural property (orthonormal rows) is
precisely what makes a matrix's transpose equal to its inverse: when
you multiply an orthonormal matrix by its own transpose, each entry
of the result is a dot product between two of the original rows - the
diagonal entries become each row's own length squared (which is 1,
since they're unit vectors), and the off-diagonal entries become dot
products between different, perpendicular rows (which are 0) - giving
exactly the identity matrix, which is the defining property of an
inverse.

WHY THIS FAILS FOR A SCALING MATRIX
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A matrix that stretches an object twice as wide as it is tall (for
example, a 2D scaling matrix like [[2,0],[0,1]]) does not have unit-
length rows - its first row, (2,0), has length 2, not 1. Since the
rows are no longer unit vectors, multiplying the matrix by its own
transpose no longer produces the identity matrix: for this example,
M * M^T = [[2,0],[0,1]] * [[2,0],[0,1]] = [[4,0],[0,1]], which is
clearly not the identity matrix. The actual inverse of this scaling
matrix is [[0.5,0],[0,1]] - the reciprocal of each scale factor - a
completely different matrix from its transpose (which, since the
matrix happens to be symmetric in this particular example, is
actually identical to the original matrix itself, not its inverse).

THE GENERAL PRINCIPLE
------------------------------
The R^-1=R^T shortcut is not a general property of all matrices - it
is specifically a consequence of a matrix's rows/columns being
orthonormal (mutually perpendicular unit vectors), which is exactly
what characterizes a pure rotation (and reflection) matrix and
nothing else. Any transformation that also scales, shears, or
otherwise distorts lengths breaks this property, and requires a
genuine matrix inversion (of the kind Numerical Methods & Floating-
Point Computation's own Chapter 8 covered) instead of the cheap
transpose shortcut.

WHY THIS WORKS AS AN ANSWER
------------------------------
The explanation identifies the precise structural property
(orthonormal rows) that makes the shortcut valid, demonstrates
concretely with a specific scaling matrix why that property fails
once scaling is introduced, and generalizes to the correct broader
principle (the shortcut applies to orthonormal transformations
specifically, not matrices in general).