Exercise 1: Solving a System by Gaussian Elimination — Possible Solution
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GIVEN
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3x + y = 11
x - y = 1

STEP 1: THE AUGMENTED MATRIX
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[3,  1 | 11]
[1, -1 |  1]

STEP 2: ELIMINATING x FROM ROW 2
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The row operation used is R2 -> R2 - (1/3) R1, since dividing row 2's
leading coefficient's target (1) by row 1's leading coefficient (3)
gives a factor of 1/3:

  New row 2, column 1: 1 - (1/3)(3)  = 0
  New row 2, column 2: -1 - (1/3)(1) = -1 - 1/3 = -4/3
  New row 2, RHS:       1 - (1/3)(11) = 1 - 11/3 = -8/3

Resulting echelon form:
[3,  1    | 11 ]
[0, -4/3  | -8/3]

STEP 3: BACK-SUBSTITUTION
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Row 2 reads: -(4/3) y = -8/3
  y = (-8/3) / (-4/3) = 2

Substituting y = 2 into row 1: 3x + 2 = 11
  3x = 9
  x = 3

FINAL ANSWER: x = 3, y = 2

STEP 4: CHECKING AGAINST BOTH ORIGINAL EQUATIONS
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3x + y = 3(3) + 2 = 9 + 2 = 11  -- matches the first equation
x - y  = 3 - 2 = 1              -- matches the second equation

WHY THIS WORKS AS AN ANSWER
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The elimination follows this chapter's own row-operation and echelon-
form procedure exactly, the back-substitution is shown as two
separate steps (solve the bottom row, then substitute upward), and
the final values are independently checked against both of the
original, unmodified equations rather than only the reduced form.