Exercise 2: Solving a System via the Inverse, Cross-Checked Against Chapter 6 — Possible Solution
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GIVEN
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A = [[3, 1], [1, -1]]
b = [11, 1]
(the system 3x + y = 11, x - y = 1, from Chapter 6's own Exercise 1)

STEP 1: THE DETERMINANT
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det(A) = (3)(-1) - (1)(1) = -3 - 1 = -4

STEP 2: COMPUTING A^-1
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With a=3, b=1, c=1, d=-1:

  Swap diagonal, negate off-diagonal: [[-1, -1], [-1, 3]]
  Multiply by 1/(-4):

  A^-1 = [[1/4, 1/4], [1/4, -3/4]]

STEP 3: SOLVING x = A^-1 b
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x (first unknown) = (1/4)(11) + (1/4)(1) = 11/4 + 1/4 = 12/4 = 3
y (second unknown) = (1/4)(11) + (-3/4)(1) = 11/4 - 3/4 = 8/4 = 2

x = A^-1 b = [3, 2]

STEP 4: CONFIRMING AGAINST CHAPTER 6
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Chapter 6's own Gaussian elimination on this exact system found x = 3,
y = 2. The inverse-based method here produces the identical answer,
[3, 2], confirming that both this chapter's inverse approach and
Chapter 6's elimination approach are solving the same underlying
problem and must always agree on a system with a unique solution.

WHY THIS WORKS AS AN ANSWER
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The inverse is built using this chapter's own formula, the
multiplication x = A^-1 b is carried out entry by entry, and the
final result is explicitly compared against the independently-
obtained Chapter 6 answer rather than left unchecked - directly
demonstrating that the two solution methods are equivalent, not just
asserting it.