Exercise 2: Rank via Row Reduction in 3D — Possible Solution
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GIVEN
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p = [1, 1, 0]
q = [0, 1, 1]
r = [1, 2, 1]

STEP 1: ROW REDUCTION
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Stack the three vectors as rows:

  [1, 1, 0]
  [0, 1, 1]
  [1, 2, 1]

R3 -> R3 - R1:

  [1, 1, 0]
  [0, 1, 1]
  [0, 1, 1]

R3 -> R3 - R2:

  [1, 1, 0]
  [0, 1, 1]
  [0, 0, 0]

STEP 2: READING OFF THE RANK
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Two rows survive as nonzero ([1,1,0] and [0,1,1]); the third reduced
completely to [0, 0, 0]. Per this chapter's own row-reduction
technique, the rank of {p, q, r} is 2, not 3 - the set is NOT
linearly independent.

STEP 3: EXPRESSING r AS A COMBINATION OF p AND q
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Since row 3 vanished after subtracting row 1 and then row 2, that is
exactly the statement r - p - q = [0, 0, 0], which rearranges to:

  r = p + q

Checking directly: p + q = [1+0, 1+1, 0+1] = [1, 2, 1] = r exactly.

WHY THIS WORKS AS AN ANSWER
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The row reduction follows this chapter's own general independence-
testing procedure step by step, the rank is read directly from the
number of surviving nonzero rows, and the redundant vector is
expressed as a combination of the other two by reading the exact row
operations used to zero it out, then independently confirmed with
direct component-by-component addition.