Exercise 1: Confirming Non-Commutativity With (0,1,0) — Possible Solution
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ORDER 1: Rx(90°) FIRST, THEN Ry(90°)
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Starting point: (0,1,0).

Step 1 - apply Rx(90°): rotating around the x-axis leaves the
x-coordinate unchanged and rotates the (y,z) pair. With cos(90)=0 and
sin(90)=1:
    Rx(90) * (0,1,0) = (0, 0*1 - 1*0, 1*1 + 0*0) = (0, 0, 1)

Step 2 - apply Ry(90°) to (0,0,1): rotating around the y-axis leaves
the y-coordinate unchanged and rotates the (x,z) pair:
    Ry(90) * (0,0,1) = (0*0 + 1*1, 0, -1*0 + 0*1) = (1, 0, 0)

Final result: (1, 0, 0)

ORDER 2: Ry(90°) FIRST, THEN Rx(90°)
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Starting point: (0,1,0).

Step 1 - apply Ry(90°): rotating (0,1,0) around the y-axis leaves
the y-coordinate (1) unchanged, and only touches x and z, both of
which start at 0:
    Ry(90) * (0,1,0) = (0, 1, 0)   (unchanged, since x=z=0 here)

Step 2 - apply Rx(90°) to (0,1,0): rotates the (y,z) pair:
    Rx(90) * (0,1,0) = (0, 0*1-1*0, 1*1+0*0) = (0, 0, 1)

Final result: (0, 0, 1)

CONFIRMING THE TWO ORDERS DIFFER
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Order 1 (Rx then Ry) produces (1, 0, 0).
Order 2 (Ry then Rx) produces (0, 0, 1).

These are two clearly different points - not just numerically close,
but pointing along entirely different coordinate axes - confirming
that applying the same two 90-degree rotations in opposite orders to
the same starting point gives genuinely different final results.

WHY THIS WORKS AS AN ANSWER
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The answer works through both rotation orders step by step using this
chapter's own rotation-matrix definitions, arriving at two explicitly
different final points, directly confirming the chapter's own claim
that 3D rotations around different axes do not commute.