MathematicsChapter 3: Matrices and Determinants

INVERSE OF A 2x2 MATRIX

If |A| != 0, A^{-1} = (1/|A|) adj(A).

Learning objectives

Explain inverse of a 2x2 matrix using the correct definitions, conditions, and notation.
Apply (1/|A|) adj(A), swap diagonal, negate off correctly in routine and exam-style questions.
Show each important step, simplify the result, and check that it satisfies the original conditions.

Definition

If |A| != 0, A^-1 = (1/|A|) adj(A). For [[a,b],[c,d]], adj(A) = [[d, -b], [-c, a]] (swap the main diagonal, negate the other two). Then A A^-1 = I. If |A|=0, stop: no inverse.

Where / symbols

INVERSE OF A 2x2 MATRIX: If |A| != 0, A^{-1} = (1/|A|) adj(A). For [[a,b],[c,d]], adj(A) = [[d, -b], [-c, a]] (swap the main diagonal, negate the other two). Then A A^{-1} = I. If |A|=0, stop: no inverse.
ADJOINT: A=[[2,3],[1,4]] has adj [[4,-3],[-1,2]]. Swap 2 with 4, flip signs of 3 and 1. Do not swap a with d and also swap b with c wrongly.
SCALE BY 1/|A|: |A|=8-3=5, so A^{-1}=(1/5)[[4,-3],[-1,2]] = [[4/5, -3/5], [-1/5, 2/5]]. Check: A times this should be I.

Overview

Det first, then adjoint:

Adjoint

A=[[2,3],[1,4]] has adj [[4,-3],[-1,2]]. Swap 2 with 4, flip signs of 3 and 1. Do not swap a with d and also swap b with c wrongly.

Examples adj[[2,3],[1,4]]=[[4,-3],[-1,2]] · Not the transpose only · Not without the minuses

Scale By 1/|a|

|A|=8-3=5, so A^-1=(15)[[4,-3],[-1,2]] = [[45, -35], [-15, 25]]. Check: A times this should be I.

Examples |A|=5 · A^{-1} exists · AA^{-1}=I

WHAT — INVERSE OF A 2x2 MATRIX

If |A| != 0, A^-1 = (1/|A|) adj(A). For [[a,b],[c,d]], adj(A) = [[d, -b], [-c, a]] (swap the main diagonal, negate the other two). Then A A^{-1} = I. If |A|=0, stop: no inverse. The symbols and conditions are part of the mathematics, so include them when stating the rule.

WHY — Why this topic matters

Inverse Of A 2X2 Matrix matters because matrices organise numbers and give short methods for transformations and simultaneous equations. Correct order, determinant, and multiplication rules prevent most exam errors.

HOW — How to solve and explain it

Start by identifying what is given and what must be found. Work through ADJOINT, then SCALE BY 1/|A|. Worked: A = [[2, 3], [1, 4]]. Find A^-1. |A| = 24 - 31 = 8 - 3 = 5 adj(A) = [[4, -3], [-1, 2]] A^-1 = (15) [[4, -3], [-1, 2]] A^-1 = [[45, -35], [-15, 25]] Finish by checking signs, restrictions, units, and whether the answer matches the question.

Worked example

A = [[2, 3], [1, 4]]. Find A^-1.
|A| = 2*4 - 3*1 = 8 - 3 = 5
adj(A) = [[4, -3], [-1, 2]]
A^-1 = (15) [[4, -3], [-1, 2]]
A^-1 = [[45, -35], [-15, 25]]

In simple words

If |A|=0 you cannot 'divide by zero' to make an inverse. Write 'does not exist'.

Key points

(1/|A|) adj(A)
swap diagonal, negate off
[[d,-b],[-c,a]]
divide every adj entry

Exam preparation

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