2.2 Matrices


2026 Syllabus Objectives

By the end of these notes, you will be able to:

  1. Write a system of 3 linear simultaneous equations as a matrix equation, and vice versa.
  2. Understand when 3 simultaneous equations have solutions (consistent) or no solutions (inconsistent), link this to whether the matrix is singular, and interpret the situation geometrically using planes.
  3. Understand the terms characteristic equation, eigenvalue, and eigenvector for square matrices, and use the definition Ae=λeA\mathbf{e} = \lambda\mathbf{e} to prove simple properties.
  4. Find eigenvalues and eigenvectors of 2×2 and 3×3 matrices (for real, distinct eigenvalues only).
  5. Write a square matrix in the form QDQ1QDQ^{-1}, where DD is a diagonal matrix of eigenvalues and QQ is the matrix of eigenvectors, and use this to calculate powers of matrices.
  6. Use the fact that every square matrix satisfies its own characteristic equation (e.g. to find powers or the inverse of a matrix).

Section 1 — Writing Simultaneous Equations as a Matrix Equation

What is a system of linear simultaneous equations?

A linear simultaneous equation is simply an equation involving unknowns (like xx, yy, zz) where none of them are squared or multiplied together — just plain multiples of each variable added together. A "system" means you have several of these equations at the same time, and you want to find values of xx, yy, zz that satisfy all of them at once.

For example, consider this system of 3 equations in 3 unknowns:

2x+yz=52x + y - z = 5 x3y+2z=4x - 3y + 2z = 4 3x+2y+z=13x + 2y + z = 1

Writing it as a matrix equation

A matrix equation takes the form:

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

Where:

  • AA is a 3×3 matrix made from the coefficients (the numbers in front of xx, yy, zz)
  • x\mathbf{x} is a column vector of the unknowns
  • b\mathbf{b} is a column vector of the right-hand side numbers

For the example above:

A=(211132321),x=(xyz),b=(541)A = \begin{pmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 3 & 2 & 1 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 5 \\ 4 \\ 1 \end{pmatrix}

So the full matrix equation is:

(211132321)(xyz)=(541)\begin{pmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \\ 1 \end{pmatrix}

You can check this works by multiplying out — each row of AA times x\mathbf{x} gives one of the original equations.

Going the other way: If you are given the matrix equation Ax=bA\mathbf{x} = \mathbf{b}, you can write out the three separate equations by reading off each row of AA.

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