Matrices

Further Maths · Further Pure Mathematics 2. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

Test this topic free All Further Maths topics

The idea

Three linear equations in x, y and z can be written as one matrix equation Ax = b. Each equation is a plane. If A is non-singular there is one solution and the planes meet at a point. If det A = 0 there is either no solution (the system is inconsistent) or infinitely many, for example when the planes share a common line.

An eigenvector of A is a non-zero vector e whose direction is unchanged by A: Ae = λe. The number λ is the eigenvalue.

Eigenvalues and eigenvectors let you write A = QDQ⁻¹ with D diagonal, which makes powers of A easy. A matrix also satisfies its own characteristic equation.

Rules to remember

det A ≠ 0: unique solution x = A⁻¹b, three planes meet at one point
det A = 0: no unique solution. Consistent means infinitely many solutions (planes meet in a line, or are the same plane); inconsistent means no common point (parallel planes or a triangular prism)
Eigenvalues: solve the characteristic equation det(A − λI) = 0. Eigenvectors: solve (A − λI)e = 0 for each λ
For a 2 × 2 matrix: λ² − (trace)λ + det A = 0, so sum of eigenvalues = trace (a + d) and product = det A
If Ae = λe then Aⁿe = λⁿe, and A⁻¹e = (1/λ)e when A is non-singular
A = QDQ⁻¹ and Aⁿ = QDⁿQ⁻¹; the columns of Q are eigenvectors in the same order as the eigenvalues on the diagonal of D
A satisfies its own characteristic equation: for 2 × 2, A² − (trace)A + (det A)I = 0

Common mistake

Students think det A = 0 always means 'no solution'. It only means there is no unique solution: test whether the equations are consistent to decide between no solution and infinitely many.

Worked example

A is the matrix with rows (5, 2) and (2, 2). Find its eigenvalues and an eigenvector for each.

Trace = 5 + 2 = 7 and det A = 5 × 2 − 2 × 2 = 6, so λ² − 7λ + 6 = 0, giving λ = 1 or λ = 6.
λ = 6: (5 − 6)x + 2y = 0, so x = 2y. Eigenvector (2, 1). Check: A(2, 1) = (10 + 2, 4 + 2) = (12, 6) = 6(2, 1).
λ = 1: (5 − 1)x + 2y = 0, so y = −2x. Eigenvector (1, −2). Check: A(1, −2) = (5 − 4, 2 − 4) = (1, −2).

Answer: λ = 6 with eigenvector (2, 1); λ = 1 with eigenvector (1, −2)

Ask about this

Type what you are stuck on. The helper finds the explanation, worked examples and the notes to read, free and instant.

Read the full notes

Matrices

Keep going

← Hyperbolic functions Differentiation →Take a Further Maths test