---
title: "Condition for matrix inverse to exist"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:condition-for-matrix-inverse-to-exist"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/condition-for-matrix-inverse-to-exist
site: BCT Engineering
---

# Condition for matrix inverse to exist

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2083 Baishakh, 5 marks; 2081 Baishakh, 5 marks; 2076 Asoj, 5 marks; 2073 Chaitra, 5 marks; 2073 Shrawan, 5 marks (asked 5 times)
- Video: [A simple condition for when the matrix inverse exists | Linear algebra makes sense](https://www.youtube.com/watch?v=ESKcF8XFzLM) — Looking Glass Universe
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/condition-for-matrix-inverse-to-exist
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Prove that the necessary and sufficient condition for a square matrix A to possess an inverse is that A is non-singular.

## Short answer

A square matrix A has an inverse iff |A| ≠ 0. Necessity: |A||A⁻¹| = 1 ⇒ |A| ≠ 0. Sufficiency: if |A| ≠ 0, A⁻¹ = adj(A)/|A| exists.

## Answer

A square matrix A of order n has an inverse if and only if |A| ≠ 0 (A is non-singular). We prove both directions.

**Tool used:** for any square matrix, A(adj A) = (adj A)A = |A| Iₙ, where adj A is the transpose of the matrix of cofactors.

#### Necessary condition (inverse exists ⇒ |A| ≠ 0)
Let A⁻¹ exist, so AA⁻¹ = I. Taking determinants and using |AB| = |A||B|:

```text
|A| · |A⁻¹| = |I| = 1
```

The product of the two numbers is 1, so neither can be zero. Hence |A| ≠ 0.

#### Sufficient condition (|A| ≠ 0 ⇒ inverse exists)
Let |A| ≠ 0. From A(adj A) = (adj A)A = |A| I, divide by the non-zero scalar |A|:

```text
A · (adj A / |A|) = (adj A / |A|) · A = I
```

So the matrix B = adj A / |A| satisfies AB = BA = I. By definition B is the inverse of A, so A⁻¹ = adj A / |A| exists.

#### Uniqueness of the inverse
If B and C are both inverses, B = BI = B(AC) = (BA)C = IC = C.

Thus A has an inverse if and only if A is non-singular. ∎

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is |A| != 0 shown when A^-1 exists?
   A. Because A^-1 = adj A, so |A| = 0 is impossible
   B. |A||A^-1| = |I| = 1, so neither determinant can be zero
   C. Because |A| + |A^-1| = 1
   D. Because |I| = 0
   Answer: B. Taking determinants of AA^-1 = I gives a product equal to 1, so |A| cannot be 0. (Key point: Necessary condition)

2. Which identity gives the inverse when |A| != 0?
   A. A(adj A) = 0, so A^-1 = adj A
   B. A + adj A = |A| I, so A^-1 = |A| adj A
   C. A(adj A) = (adj A)A = |A| I, so A^-1 = adj A / |A|
   D. A(adj A) = I, so A^-1 = adj A
   Answer: C. Dividing A(adj A) = |A| I by the non-zero scalar |A| shows adj A/|A| is the inverse. (Key point: Sufficient condition)

3. How is it shown that the inverse is unique?
   A. If B and C are inverses, B = B(AC) = (BA)C = C
   B. Because the determinant of the inverse is 1
   C. Because adj A is unique, B must equal C only if A is symmetric
   D. Because B + C = 2I
   Answer: A. Associativity with BA = I and AC = I gives B = C. (Key point: Uniqueness of inverse)

## Close variants set in other exams

- [Condition for matrix inverse to exist](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Acondition-for-matrix-inverse-to-exist-two) — Mathematics III, chapter 1, asked 6×