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Solving Linear Systems

🧮 Solving Linear Systems with the Gauss–Jordan Method​

A linear system of equations can be written in matrix form as:

Ax=bA \mathbf{x} = \mathbf{b}

where

A=[a11a12a13a21a22a23a31a32a33],x=[x1x2x3],b=[b1b2b3]A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \\[4pt] x_2 \\[4pt] x_3 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} b_1 \\[4pt] b_2 \\[4pt] b_3 \end{bmatrix}

The Gauss–Jordan method transforms the matrix AA into the identity matrix by using row operations, producing the solution vector x\mathbf{x}.

In practice, we can use numerical libraries (like NumPy) that implement efficient algorithms based on LU/Gaussian elimination.

🔹 Example in Python​

import numpy as np

# Coefficient matrix (3x3)
A = np.array([[2, 1, -1],
              [-3, -1, 2],
              [-2, 1, 2]], dtype=float)

# Constants vector (1D is common for np.linalg.solve)
b = np.array([8, -11, -3], dtype=float)

# Solve the system A * x = b
x = np.linalg.solve(A, b)

print("Solution:")
print(x)

Output:

[ 2.  3. -1.]

📘 Mathematical Explanation​

We are solving the system:

{2x+y−z=8−3x−y+2z=−11−2x+y+2z=−3\begin{cases} 2x + y - z = 8 \\ -3x - y + 2z = -11 \\ -2x + y + 2z = -3 \end{cases}

By applying Gauss–Jordan elimination we reduce the augmented matrix

[21−18−3−12−11−212−3]\left[ \begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array} \right]

to the identity matrix on the left:

[10020103001−1]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1 \end{array} \right]

Therefore, the solution is:

x=2,y=3,z=−1x = 2, \quad y = 3, \quad z = -1