Solving Ax = b

Solving a Linear System

A system of linear equations like:

$2x + y = 5$ $x + y = 3$

can be written in matrix form as $mathbf{A}mathbf{x} = mathbf{b}$:

A = [[2, 1],
     [1, 1]]
b = [5, 3]

Gaussian Elimination

We solve this by row reduction — transforming $mathbf{A}$ into upper triangular form, then back-substituting:

def solve(A, b):
    n = len(b)
    # Augment [A|b]
    M = [A[i][:] + [b[i]] for i in range(n)]
    # Forward elimination
    for col in range(n):
        for row in range(col+1, n):
            factor = M[row][col] / M[col][col]
            M[row] = [M[row][k] - factor*M[col][k] for k in range(n+1)]
    # Back substitution
    x = [0.0] * n
    for i in range(n-1, -1, -1):
        x[i] = (M[i][n] - sum(M[i][j]*x[j] for j in range(i+1, n))) / M[i][i]
    return [round(xi, 1) for xi in x]

x = solve([[2,1],[1,1]], [5,3])
print(x)  # [2.0, 1.0]  →  x=2, y=1

Why Gaussian Elimination?

• More numerically stable than computing the inverse
• Works directly on the augmented matrix $[mathbf{A} mid mathbf{b}]$
• Basis for LU decomposition

Verifying the Solution

Check: $mathbf{A}mathbf{x}$ should equal $mathbf{b}$:

Ax = [sum(A[i][j]*x[j] for j in range(len(x))) for i in range(len(A))]
print(all(abs(Ax[i] - b[i]) < 1e-9 for i in range(len(b))))  # True

Your Task

Implement solve(A, b) that returns the solution vector $mathbf{x}$ to $mathbf{A}mathbf{x} = mathbf{b}$, rounded to 1 decimal place.