10 min readConcept sheet1-mark MCQ and assertion-reason; 3-mark questions; the first half of the 5-mark matrix-method question

Inverse of a square matrix

A square matrix A has an inverse exactly when |A| ≠ 0. Then A⁻¹ = (1/|A|) adj A, and AA⁻¹ = A⁻¹A = I.

Inverse formulaA⁻¹ = (1/|A|) adj A, when |A| ≠ 0

When it existsA is invertible exactly when A is non-singular, that is |A| ≠ 0

Products|AB| = |A||B| and (AB)⁻¹ = B⁻¹A⁻¹ (note the reversed order)

Determinant of the inverse|A⁻¹| = 1/|A|

What the inverse is

If AB = BA = I, then B is the inverse of A, written A⁻¹. You saw in the Matrices chapter that the inverse, when it exists, is unique. Start from A(adj A) = |A| I and divide both sides by |A|. This is allowed only when |A| ≠ 0, and it gives A⁻¹ = (1/|A|) adj A.

How to find A⁻¹

1★
Find |A|. If |A| = 0, stop: A is singular and has no inverse.
2
Find all the cofactors Aᵢⱼ.
3
Write adj A, the transpose of the cofactor matrix.
4
Divide every element of adj A by |A|: A⁻¹ = (1/|A|) adj A.
5
Check one row: A × A⁻¹ should give the matching row of I.

★ key step

Example 1: a 2 × 2 matrix

Problem

Find the inverse of A = [3, −1; 4, 2].

Working

|A| = 3 × 2 − (−1) × 4 = 6 + 4 = 10 ≠ 0, so A⁻¹ exists.
adj A = [2, 1; −4, 3] (swap 3 and 2, change the signs of −1 and 4).
A⁻¹ = (1/10)[2, 1; −4, 3]
Check: A × adj A = [6 + 4, 3 − 3; 8 − 8, 4 + 6] = [10, 0; 0, 10] = 10 I.

Answer

A⁻¹ = (1/10)[2, 1; −4, 3]

Example 2: a 3 × 3 matrix

Problem

Find A⁻¹ for A = [1, 2, 1; 0, 1, −1; 2, 1, 3].

Working

|A| = 1(3 + 1) − 2(0 + 2) + 1(0 − 2) = 4 − 4 − 2 = −2 ≠ 0
Row 1 cofactors: A₁₁ = 4, A₁₂ = −2, A₁₃ = −2
Row 2 cofactors: A₂₁ = −5, A₂₂ = 1, A₂₃ = 3
Row 3 cofactors: A₃₁ = −3, A₃₂ = 1, A₃₃ = 1
adj A = [4, −5, −3; −2, 1, 1; −2, 3, 1]
A⁻¹ = (1/|A|) adj A = (−1/2)[4, −5, −3; −2, 1, 1; −2, 3, 1]

Answer

A⁻¹ = (−1/2)[4, −5, −3; −2, 1, 1; −2, 3, 1], that is [−2, 5/2, 3/2; 1, −1/2, −1/2; 1, −3/2, −1/2]

Example 3: inverse from a matrix equation

Problem

For A = [2, 3; 1, 4], show that A² − 6A + 5I = O. Hence find A⁻¹.

Working

A² = [2, 3; 1, 4][2, 3; 1, 4] = [4 + 3, 6 + 12; 2 + 4, 3 + 16] = [7, 18; 6, 19]
6A = [12, 18; 6, 24] and 5I = [5, 0; 0, 5]
A² − 6A + 5I = [7 − 12 + 5, 18 − 18 + 0; 6 − 6 + 0, 19 − 24 + 5] = [0, 0; 0, 0] = O
|A| = 8 − 3 = 5 ≠ 0, so A⁻¹ exists. Multiply A² − 6A + 5I = O by A⁻¹.
A − 6I + 5A⁻¹ = O, so A⁻¹ = (1/5)(6I − A)
A⁻¹ = (1/5)[6 − 2, 0 − 3; 0 − 1, 6 − 4] = (1/5)[4, −3; −1, 2]

Answer

A⁻¹ = (1/5)[4, −3; −1, 2]

Singular vs non-singular

Singular matrix|A| = 0. No inverse. A(adj A) = O.

Non-singular matrix|A| ≠ 0. The inverse exists and is unique: A⁻¹ = (1/|A|) adj A.

Always find |A| first. It decides whether the rest of the work is needed.

In real life

Simple secret codes turn letters into numbers and multiply them by a matrix. The receiver multiplies by the inverse matrix to read the message, so the coding matrix must have |A| ≠ 0.

Exam angle

1 marks: “If A is an invertible matrix of order 3 and |A| = 5, find |A⁻¹|.” AA⁻¹ = I, so |A||A⁻¹| = 1. |A⁻¹| = 1/5.

1 marks: “For what value of k does A = [k, 6; 2, 3] have no inverse?” No inverse when |A| = 0: 3k − 12 = 0, so k = 4.

3 marks: “If A = [2, −1; −1, 2], show that A² − 4A + 3I = O. Hence find A⁻¹.” A² = [5, −4; −4, 5] and 4A = [8, −4; −4, 8], so A² − 4A + 3I = O. Multiply by A⁻¹: A − 4I + 3A⁻¹ = O, so A⁻¹ = (1/3)(4I − A) = (1/3)[2, 1; 1, 2].

Remember it

No determinant, no inverse. A⁻¹ = adj A over |A|, and (AB)⁻¹ reverses the order: B⁻¹A⁻¹.