Exercise 2: Composing S and R in Both Orders — Possible Solution
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GIVEN
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w = [2, -1]
S = [[3, 0], [0, 2]]
R = [[0, -1], [1, 0]]

STEP 1: COMPUTING R x S ("SCALE FIRST, THEN ROTATE")
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Row 0, Col 0: (0)(3) + (-1)(0) = 0
Row 0, Col 1: (0)(0) + (-1)(2) = -2
Row 1, Col 0: (1)(3) + (0)(0) = 3
Row 1, Col 1: (1)(0) + (0)(2) = 0

R x S = [[0, -2], [3, 0]]

Applying to w:
  (R x S) w = [(0)(2) + (-2)(-1), (3)(2) + (0)(-1)] = [0 + 2, 6 + 0] = [2, 6]

STEP 2: COMPUTING S x R ("ROTATE FIRST, THEN SCALE")
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Row 0, Col 0: (3)(0) + (0)(1) = 0
Row 0, Col 1: (3)(-1) + (0)(0) = -3
Row 1, Col 0: (0)(0) + (2)(1) = 2
Row 1, Col 1: (0)(-1) + (2)(0) = 0

S x R = [[0, -3], [2, 0]]

Applying to w:
  (S x R) w = [(0)(2) + (-3)(-1), (2)(2) + (0)(-1)] = [0 + 3, 4 + 0] = [3, 4]

STEP 3: CONFIRMING THE RESULTS DIFFER
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(R x S) w = [2, 6]
(S x R) w = [3, 4]

The two results are different, confirming order genuinely matters
here, not just in principle.

WHICH ORDER MEANS WHAT
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R x S applied to w = [2, 6] represents "scale w first, then rotate the
scaled result" (S is the rightmost matrix, applied first).
S x R applied to w = [3, 4] represents "rotate w first, then scale the
rotated result" (R is the rightmost matrix, applied first).

WHY THIS WORKS AS AN ANSWER
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Both combined matrices are computed using this chapter's own row-by-
column multiplication rule, applied to the same starting vector w, and
the "which order means what" explanation follows directly from this
chapter's own stated convention that A x B means "apply B first, then
A" - rather than guessing which combined matrix corresponds to which
real-world sequence.