Solving Systems of Equations with Matrices Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Solving Systems of Equations with Matrices.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

Systems of linear equations can be represented as the matrix equation Ax = b and solved using augmented matrices with row reduction (Gaussian elimination), matrix inverses (x = A^{-1}b), or Cramer's rule (using determinants).

Instead of juggling multiple equations with substitution or elimination, pack everything into a matrix and use systematic row operations. It is like organizing a messy desk—once the equations are neatly arranged in a matrix, a mechanical process (row reduction) reveals the answer. Each row operation is an allowed algebraic move (swap equations, scale an equation, add equations) performed on the matrix.

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How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Matrices provide a systematic, algorithmic approach to solving systems of equations that scales to any number of variables.

Common stuck point: Row reduction errors accumulate. Track each step carefully and double-check arithmetic. When using Cramer's rule, remember each variable's determinant replaces a different column of the coefficient matrix.

Sense of Study hint: Write the augmented matrix carefully, double-checking each coefficient's position, before starting row operations.

Worked Examples

Example 1

medium
Solve using the inverse matrix: \begin{cases} 2x + y = 5 \\ x - y = 1 \end{cases}

Solution

  1. 1
    Step 1: Write as Ax = b: \begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \end{bmatrix}.
  2. 2
    Step 2: \det(A) = -2 - 1 = -3. A^{-1} = \frac{1}{-3}\begin{bmatrix} -1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1/3 & 1/3 \\ 1/3 & -2/3 \end{bmatrix}.
  3. 3
    Step 3: x = A^{-1}b = \begin{bmatrix} 1/3 & 1/3 \\ 1/3 & -2/3 \end{bmatrix}\begin{bmatrix} 5 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}.
  4. 4
    Check: 2(2) + 1 = 5 ✓ and 2 - 1 = 1 ✓

Answer

x = 2, y = 1
Any system Ax = b with invertible A can be solved as x = A^{-1}b. This matrix method is equivalent to elimination but becomes more systematic for larger systems.

Example 2

hard
Use Cramer's rule to solve: \begin{cases} 3x + 2y = 7 \\ x - y = 1 \end{cases}

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Write the system \begin{cases} x + 3y = 7 \\ 2x - y = 4 \end{cases} as a matrix equation Ax = b.

Example 2

medium
Solve using the inverse: \begin{cases} x + 2y = 8 \\ 3x + 5y = 19 \end{cases}
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Background Knowledge

These ideas may be useful before you work through the harder examples.

systems of equationsinverse matrixdeterminant