---
title: "Property A(adj A) = |A|I"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:property-a-adj-a-a-i"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/property-a-adj-a-a-i
site: BCT Engineering
---

# Property A(adj A) = |A|I

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2075 Chaitra, 5 marks; 2074 Chaitra, 5 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/property-a-adj-a-a-i
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

If A is a square matrix of order n, prove that A(adj A) = (adj A)A = |A| Iₙ, where Iₙ is a unit matrix having the same order as A.

## Short answer

Using cofactor expansion, the diagonal entries of A(adj A) equal |A| and off-diagonal entries are zero, giving |A| I. The same holds for (adj A)A, so A(adj A) = (adj A)A = |A| I.

## Answer

**Definitions.** Let A = [aᵢⱼ] be of order n and Aᵢⱼ the cofactor of aᵢⱼ. Then adj A = [Aᵢⱼ]ᵀ, so the (i, j) entry of adj A is Aⱼᵢ.

#### Proof that A(adj A) = |A| I
The (i, j) entry of A(adj A) is

```text
Σₖ aᵢₖ · Aⱼₖ      (k = 1 to n)
```

Two cases are used from the properties of determinants:

1. **i = j:** Σₖ aᵢₖ Aᵢₖ is the expansion of |A| along row i, so it equals |A|.
2. **i ≠ j:** Σₖ aᵢₖ Aⱼₖ is the expansion along row j of the determinant obtained by replacing row j with row i. That determinant has two identical rows, so its value is 0.

Hence all diagonal entries are |A| and all off-diagonal entries are 0:

```text
A(adj A) = [|A|  0  …  0 ]
           [ 0  |A| …  0 ] = |A| Iₙ
           [ …          … ]
           [ 0   0  … |A|]
```

#### Proof that (adj A)A = |A| I
The (i, j) entry of (adj A)A is Σₖ Aₖᵢ aₖⱼ.

- i = j: this is the expansion of |A| along column i, so it equals |A|.
- i ≠ j: this is the expansion along column i of the determinant obtained by replacing column i with column j. That determinant has two identical columns, so it equals 0.

So (adj A)A = |A| Iₙ also.

**Therefore A(adj A) = (adj A)A = |A| Iₙ.** ∎

**Consequence:** if |A| ≠ 0, then A⁻¹ = adj A / |A|.

## Check yourself

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

1. adj A is
   A. the transpose of the cofactor matrix
   B. the cofactor matrix without transposing
   C. the inverse of A
   D. the matrix of minors only
   Answer: A. adj A = [A_ij]^T. (Key point: Adjoint definition)

2. Diagonal entries of A(adj A) equal |A| because
   A. each a_ik equals 1
   B. sum a_ik A_ik is the expansion along row i
   C. adj A is the identity
   D. the cofactors are zero
   Answer: B. Row expansion gives the determinant. (Key point: Diagonal entries)

3. Off-diagonal entries are zero because
   A. the cofactors are negative
   B. A is symmetric
   C. the sum is a determinant with two identical rows
   D. A is singular
   Answer: C. Replacing row j with row i gives equal rows, so 0. (Key point: Off-diagonal entries)

4. If |A| != 0 then
   A. A^-1 = |A| adj A
   B. A^-1 = adj A
   C. A^-1 = A^T
   D. A^-1 = adj A / |A|
   Answer: D. Divide the identity by |A|. (Key point: Consequence)

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