Tools

System of Linear Equations

A system of linear equations consists of multiple linear equations involving the same set of variables. Solving such systems helps determine the values of unknowns that satisfy all equations simultaneously.

Consider a system of \( m \) equations with \( n \) unknowns, represented in matrix form as:

\( A\vec{x} = \vec{b} \)

where:

  • \( A \) is an \( m \times n \) coefficient matrix
  • \( \vec{x} \) is an \( n \times 1 \) vector of unknowns
  • \( \vec{b} \) is an \( m \times 1 \) constant vector

Example:

Suppose we have the following system:

\[ \begin{aligned} x + 2y &= 5 \\ 3x + 4y &= 11 \end{aligned} \]

Step-by-Step Solution Using Inverse Matrix

Step 1: Write the system in matrix form.

\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad \vec{x} = \begin{bmatrix} x \\ y \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} 5 \\ 11 \end{bmatrix}, \quad Ax = b \]

Step 2: Compute the inverse of matrix \( A \).

\[ A^{-1} = \frac{1}{(1)(4) - (2)(3)} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \]

Step 3: Multiply \( A^{-1} \) by \( b \) to find \( x \).

\[ \vec{x} = A^{-1} \vec{b} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \begin{bmatrix} 5 \\ 11 \end{bmatrix} = \begin{bmatrix} (-2)(5) + (1)(11) \\ (1.5)(5) + (-0.5)(11) \end{bmatrix} = \begin{bmatrix} -10 + 11 \\ 7.5 - 5.5 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \]

Final Solution:

\[ x = 1, \quad y = 2 \]

Interpreting Solutions:

  • Unique Solution: The system has exactly one solution if \( A \) is invertible (determinant ≠ 0).
  • Infinite Solutions: Occurs when \( A \) is singular (determinant = 0) and equations are consistent.
  • No Solution: Occurs when \( A \) is singular and equations are inconsistent.

Using the inverse matrix method is a systematic way to solve linear systems efficiently, especially for small systems or when using computational tools.