Matrices and Simultaneous Equations

Solve systems of linear equations efficiently using matrix notation

Introduction to Matrix Equations

Matrices provide a powerful way to represent and solve systems of simultaneous equations. Instead of writing multiple equations separately, we can express an entire system as a single matrix equation, making it easier to understand and solve.

This approach is particularly useful for:

  • Systems with many equations and unknowns
  • Using computer algorithms to find solutions
  • Understanding linear transformations
  • Applying consistent notation across engineering and science
Key Benefit:

Converting equations to matrix form allows us to use powerful mathematical tools like matrix inversion to solve the system directly.

Converting Equations to Matrix Form

The General Pattern

Matrix Equation Form:

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

Matrix Equation:
\(\displaystyle A \mathbf{x} = \mathbf{b}\)

Where:

  • \(A\) is the coefficient matrix (contains all coefficients)
  • \(\mathbf{x}\) is the variable vector (unknowns we're solving for)
  • \(\mathbf{b}\) is the constant vector (right-hand side values)

Concrete Example

Convert System to Matrix Form

Original system of equations:

\(\displaystyle \begin{align}x + y &= 3 \\ 2x + 3y &= 1\end{align}\)

Coefficient matrix A:

\(\displaystyle A = \begin{bmatrix}1 & 1 \\ 2 & 3\end{bmatrix}\)

Variable vector x:

\(\displaystyle \mathbf{x} = \begin{bmatrix}x \\ y\end{bmatrix}\)

Constant vector b:

\(\displaystyle \mathbf{b} = \begin{bmatrix}3 \\ 1\end{bmatrix}\)

Matrix equation:

\(\displaystyle \begin{bmatrix}1 & 1 \\ 2 & 3\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}3 \\ 1\end{bmatrix}\)

Verification by multiplication:

  • Row 1: \(1 \cdot x + 1 \cdot y = 3\) → \(x + y = 3\) ✓
  • Row 2: \(2 \cdot x + 3 \cdot y = 1\) → \(2x + 3y = 1\) ✓

Solving Using Matrix Inversion

The Solution Method

Solution Formula:

For the matrix equation \(A\mathbf{x} = \mathbf{b}\), the solution is:

Solution by Matrix Inversion:
\(\displaystyle \mathbf{x} = A^{-1}\mathbf{b}\)

This approach is directly analogous to solving a simple algebraic equation:

Analogy with Arithmetic

Simple equation: \(ax = b\) → \(x = a^{-1}b = \frac{b}{a}\)

Matrix equation: \(A\mathbf{x} = \mathbf{b}\) → \(\mathbf{x} = A^{-1}\mathbf{b}\)

Important Condition:

Matrix inversion only works if the coefficient matrix \(A\) is non-singular (i.e., its determinant is not zero). If \(A\) is singular, the system has no unique solution.

Solution Process Overview

Equations
→
Matrix Form
→
Find \(A^{-1}\)
→
Solution

Complete Worked Example

Solve: \(x + y = 3\) and \(2x + 3y = 1\)

Step 1: Write in matrix form

\(\displaystyle \begin{bmatrix}1 & 1 \\ 2 & 3\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}3 \\ 1\end{bmatrix}\)

Step 2: Calculate the inverse of A

\(\displaystyle A^{-1} = \frac{1}{(1)(3) - (1)(2)} \begin{bmatrix}3 & -1 \\ -2 & 1\end{bmatrix}\)\( = \frac{1}{1} \begin{bmatrix}3 & -1 \\ -2 & 1\end{bmatrix} = \begin{bmatrix}3 & -1 \\ -2 & 1\end{bmatrix}\)

Step 3: Multiply both sides by \(A^{-1}\)

\(\displaystyle \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}3 & -1 \\ -2 & 1\end{bmatrix} \begin{bmatrix}3 \\ 1\end{bmatrix}\)

Step 4: Perform the matrix multiplication

\(\displaystyle \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}(3)(3) + (-1)(1) \\ (-2)(3) + (1)(1)\end{bmatrix} = \begin{bmatrix}9 - 1 \\ -6 + 1\end{bmatrix} = \begin{bmatrix}8 \\ -5\end{bmatrix}\)

Solution: \(x = 8\) and \(y = -5\)

Verification:

  • Equation 1: \(8 + (-5) = 3\) ✓
  • Equation 2: \(2(8) + 3(-5) = 16 - 15 = 1\) ✓

Advantages of Matrix Method

Systematic Approach
Consistent method for any size system of equations
Computer Friendly
Easily implemented in software and programming languages
Scalability
Works equally well for 2×2 or 1000×1000 systems
Compact Notation
Single equation replaces multiple equations

When Matrix Inversion Doesn't Work

The matrix inversion method only works when the coefficient matrix \(A\) is invertible:

Scenario Determinant Inverse Exists? Solution
Unique Solution \(\det(A) \neq 0\) Yes \(\mathbf{x} = A^{-1}\mathbf{b}\)
No Solution \(\det(A) = 0\) No Equations are inconsistent
Infinite Solutions \(\det(A) = 0\) No Equations are dependent
Singular Matrices:

If \(\det(A) = 0\), you must use alternative methods like Gaussian elimination to determine if solutions exist.

Key Points to Remember

  • System of equations can be written as \(A\mathbf{x} = \mathbf{b}\)
  • \(A\) is the coefficient matrix, \(\mathbf{x}\) is variables, \(\mathbf{b}\) is constants
  • Solution exists as \(\mathbf{x} = A^{-1}\mathbf{b}\) if \(A\) is invertible
  • Matrix method is analogous to dividing both sides by \(A\)
  • Coefficient matrix must be square and non-singular
  • If \(\det(A) = 0\), the system may have no solution or infinite solutions
  • Matrix method is efficient for computer implementation
  • Works for systems of any size (2×2, 3×3, etc.)
  • Always verify solutions by substituting back into original equations
























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