Adjoint and inverse

Building the cofactor matrix, transposing it to get the adjoint, and dividing by the determinant to get the inverse — plus the Gauss–Jordan route and the properties of A⁻¹.

Module contents

Division is not defined for matrices. The inverse is what replaces it.

Definition

For a square matrix $A$ of order $n$ , the inverse $A^{- 1}$ is the matrix satisfying

$A A^{- 1} = A^{- 1} A = I_{n}$

Existence. $A^{- 1}$ exists if and only if $∣ A ∣ \neq = 0$ — that is, if and only if $A$ is non-singular. When it exists it is unique.

Route 1 — The adjoint method

Three steps, in this order. Do not skip the transpose.

Step 1 — Cofactor matrix

Replace each element $a_{ij}$ by its cofactor $A_{ij} = (- 1)^{i + j} M_{ij}$ .

$C = [A_{ij}]$

For a $3 \times 3$ this means nine $2 \times 2$ determinants and nine signs.

Step 2 — Adjoint

The adjoint (or adjugate) is the transpose of the cofactor matrix:

$adj A = C^{T} = [A_{j i}]$

$adj A = A_{11} A_{12} A_{13} A_{21} A_{22} A_{23} A_{31} A_{32} A_{33}$

Note the index order: the adjoint's $(i, j)$ entry is the cofactor $A_{j i}$ . Forgetting this transpose is the single most common error in the topic.

Step 3 — Divide by the determinant

$A^{- 1} = \frac{1}{∣ A ∣} adj A$

The shortcut for 2 × 2

For $A = [a c b d]$ :

$adj A = [d - c - b a] A^{- 1} = \frac{1}{a d - b c} [d - c - b a]$

Swap the diagonal, negate the off-diagonal, divide by $∣ A ∣$ . Memorise this; it appears constantly.

The identity behind the formula

$A (adj A) = (adj A) A = ∣ A ∣ I$

This holds for every square matrix, singular or not. Dividing through by $∣ A ∣$ — legal only when $∣ A ∣ \neq = 0$ — produces the inverse formula. Whenever a question asks you to "verify $A (adj A) = ∣ A ∣ I$ ", this is what it wants.

Route 2 — Gauss–Jordan (elementary row operations)

Often faster for $3 \times 3$ and much faster beyond, and it needs no cofactors.

Write the augmented array $[A ∣ I]$ .
Apply elementary row operations to the whole array until the left block becomes $I$ .
The right block is now $A^{- 1}$ :

$[A ∣ I] row operations [I ∣ A^{- 1}]$

If a row of the left block ever becomes all zeros, $A$ is singular and has no inverse — stop there.

The three permitted elementary row operations are:

$R_{i} \leftrightarrow R_{j}$ — interchange two rows
$R_{i} \to k R_{i}$ with $k \neq = 0$ — scale a row
$R_{i} \to R_{i} + k R_{j}$ — add a multiple of another row

Properties of the inverse

Property	Statement
Involution	$(A^{- 1})^{- 1} = A$
Reversal	$(A B)^{- 1} = B^{- 1} A^{- 1}$
Transpose	$(A^{T})^{- 1} = (A^{- 1})^{T}$
Determinant	$A^{- 1} = \frac{1}{∣ A ∣}$
Scalar	$(k A)^{- 1} = \frac{1}{k} A^{- 1}$ , $k \neq = 0$
Powers	$(A^{n})^{- 1} = (A^{- 1})^{n}$

Note that $(A B)^{- 1}$ reverses the order, exactly as the transpose does.

Properties of the adjoint

For a non-singular $A$ of order $n$ :

$∣ adj A ∣ = ∣ A ∣^{n - 1} adj (A B) = (adj B) (adj A) adj (A^{T}) = (adj A)^{T}$

$adj (adj A) = ∣ A ∣^{n - 2} A adj (k A) = k^{n - 1} adj A$

Which route to use

Case	Route
$2 \times 2$	The swap-and-negate shortcut
$3 \times 3$ , question says "using adjoint"	Adjoint method
$3 \times 3$ , method free	Either; Gauss–Jordan has fewer sign traps
$4 \times 4$ and above	Gauss–Jordan
Orthogonal matrix	$A^{- 1} = A^{T}$ , no work needed
Involutory matrix	$A^{- 1} = A$ , no work needed

Worked examples

3 solved

Every step is shown, in the order you would write it in an answer book.

Example 1

Inverse of a 2 × 2 matrix

Question

Find $A^{- 1}$ for $A = [4276]$ and verify.

Determinant
$∣ A ∣ = 4 (6) - 7 (2) = 24 - 14 = 10 \neq = 0$

Non-singular, so the inverse exists.
Adjoint by swap-and-negate
Swap $4$ and $6$ ; negate $7$ and $2$ :

$adj A = [6 - 2 - 7 4]$
Divide
$A^{- 1} = \frac{1}{10} [6 - 2 - 7 4] = [0.6 - 0.2 - 0.7 0.4]$
Verify $A A^{- 1} = I$
$A A^{- 1} = \frac{1}{10} [4276] [6 - 2 - 7 4] = \frac{1}{10} [24 - 14 12 - 12 - 28 + 28 - 14 + 24] = \frac{1}{10} [100010] = I ✓$

Answer

$A^{- 1} = \frac{1}{10} [6 - 2 - 7 4]$

Example 2

Inverse of a 3 × 3 by the adjoint method

Question

Find $A^{- 1}$ for $A = 105216340$ .

Determinant — expand along row 2 (it has a zero)
Row 2 is $(0, 1, 4)$ with signs $(-, +, -)$ :

$∣ A ∣ = - 0 + 11530 - 41526 = 1 (0 - 15) - 4 (6 - 10) = - 15 + 16 = 1$

$∣ A ∣ = 1 \neq = 0$ , so $A^{- 1}$ exists — and the division at the end will be free.
Cofactors, row 1
$A_{11} = + 1640 = 0 - 24 = - 24$
$A_{12} = - 0540 = - (0 - 20) = 20$
$A_{13} = + 0516 = 0 - 5 = - 5$
Cofactors, row 2
$A_{21} = - 2630 = - (0 - 18) = 18$
$A_{22} = + 1530 = 0 - 15 = - 15$
$A_{23} = - 1526 = - (6 - 10) = 4$
Cofactors, row 3
$A_{31} = + 2134 = 8 - 3 = 5$
$A_{32} = - 1034 = - (4 - 0) = - 4$
$A_{33} = + 1021 = 1 - 0 = 1$
Cofactor matrix, then transpose it
$C = - 24 185 20 - 15 - 4 - 5 41 adj A = C^{T} = - 24 20 - 5 18 - 15 4 5 - 4 1$
Divide by $∣ A ∣ = 1$
$A^{- 1} = \frac{1}{1} - 24 20 - 5 18 - 15 4 5 - 4 1 = - 24 20 - 5 18 - 15 4 5 - 4 1$
Verify one row
Row 1 of $A$ against column 1 of $A^{- 1}$ :

$1 (- 24) + 2 (20) + 3 (- 5) = - 24 + 40 - 15 = 1 ✓$

Row 1 of $A$ against column 2 of $A^{- 1}$ :

$1 (18) + 2 (- 15) + 3 (4) = 18 - 30 + 12 = 0 ✓$

Answer

$A^{- 1} = - 24 20 - 5 18 - 15 4 5 - 4 1$

NoteLay the nine cofactors out in a grid as you compute them, then transpose once at the end. Trying to write the adjoint directly is where sign errors creep in.

Example 3

The same inverse by Gauss–Jordan

Question

Find $A^{- 1}$ for $A = [2714]$ using elementary row operations.

Set up the augmented array
$[27141001]$
Make a leading 1 — $R_{1} \to \frac{1}{2} R_{1}$
$[17 \frac{1}{2} 4 \frac{1}{2} 0 01]$
Clear below — $R_{2} \to R_{2} - 7 R_{1}$
$[10 \frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{7}{2} 01]$
Make the second leading 1 — $R_{2} \to 2 R_{2}$
$[10 \frac{1}{2} 1 \frac{1}{2} - 7 02]$
Clear above — $R_{1} \to R_{1} - \frac{1}{2} R_{2}$
$[1001 \frac{1}{2} + \frac{7}{2} - 7 - 1 2] = [1001 4 - 7 - 1 2]$
Read off the right block
$A^{- 1} = [4 - 7 - 1 2]$

Cross-check with the shortcut: $∣ A ∣ = 8 - 7 = 1$ , and swap-and-negate gives $[4 - 7 - 1 2]$ ✓

Answer

$A^{- 1} = [4 - 7 - 1 2]$

Practice problems

7 with solutions

Work each one on paper first. The full solution — not just the answer — is one click away.

Problem 1Basic

Find the inverse of $A = [3152]$ , or show it does not exist.

Show solution

$∣ A ∣ = 3 (2) - 5 (1) = 6 - 5 = 1 \neq = 0$

Non-singular, so the inverse exists.

$adj A = [2 - 1 - 5 3]$

$A^{- 1} = \frac{1}{1} [2 - 1 - 5 3] = [2 - 1 - 5 3]$

Check: $[3152] [2 - 1 - 5 3] = [6 - 5 2 - 2 - 15 + 15 - 5 + 6] = I$ ✓

Answer

A^{- 1} = [2 - 1 - 5 3]

Problem 2Basic

Show that $A = [2163]$ has no inverse.

Show solution

$∣ A ∣ = 2 (3) - 6 (1) = 6 - 6 = 0$

Since $∣ A ∣ = 0$ , the matrix is singular and $A^{- 1}$ does not exist.

The reason is visible in the matrix: row 1 is exactly $2 \times$ row 2, so the rows are proportional and property P6 forces the determinant to zero.

Answer

∣ A ∣ = 0

, so

A

is singular and has no inverse.

Problem 3Exam level

Find $adj A$ and $A^{- 1}$ for $A = 112 12 - 1 1 - 3 3$ .

Show solution

Determinant (row 1):

$∣ A ∣ = 1 (6 - 3) - 1 (3 + 6) + 1 (- 1 - 4) = 3 - 9 - 5 = - 11 \neq = 0$

Cofactors.

$A_{11} = + 2 - 1 - 3 3 = 6 - 3 = 3 A_{12} = - 12 - 3 3 = - (3 + 6) = - 9$
$A_{13} = + 12 2 - 1 = - 1 - 4 = - 5$

$A_{21} = - 1 - 1 13 = - (3 + 1) = - 4 A_{22} = + 1213 = 3 - 2 = 1$
$A_{23} = - 12 1 - 1 = - (- 1 - 2) = 3$

$A_{31} = + 12 1 - 3 = - 3 - 2 = - 5 A_{32} = - 11 1 - 3 = - (- 3 - 1) = 4$
$A_{33} = + 1112 = 2 - 1 = 1$

Cofactor matrix and adjoint.

$C = 3 - 4 - 5 - 9 14 - 5 31 adj A = C^{T} = 3 - 9 - 5 - 4 13 - 5 41$

Inverse.

$A^{- 1} = - \frac{1}{11} 3 - 9 - 5 - 4 13 - 5 41 = \frac{1}{11} - 3 95 4 - 1 - 3 5 - 4 - 1$

Check (row 1 of $A$ × column 1 of $adj A$ ): $1 (3) + 1 (- 9) + 1 (- 5) = - 11 = ∣ A ∣$ ✓

Answer

adj A = 3 - 9 - 5 - 4 13 - 5 41

A^{- 1} = \frac{1}{11} - 3 95 4 - 1 - 3 5 - 4 - 1

Problem 4Exam level

Find $A^{- 1}$ for $A = 124 0 - 1 1 238$ by Gauss–Jordan.

Show solution

$124 0 - 1 1 238100010001$

$R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - 4 R_{1}$ :

$100 0 - 1 1 2 - 1 0 1 - 2 - 4 010001$

$R_{2} \to - R_{2}$ :

$100011210 12 - 4 0 - 1 0 001$

$R_{3} \to R_{3} - R_{2}$ :

$100010 21 - 1 12 - 6 0 - 1 1 001$

$R_{3} \to - R_{3}$ :

$100010211126 0 - 1 - 1 00 - 1$

$R_{1} \to R_{1} - 2 R_{3}$ , $R_{2} \to R_{2} - R_{3}$ :

$100010001 - 11 - 4 6 20 - 1 21 - 1$

$A^{- 1} = - 11 - 4 6 20 - 1 21 - 1$

Check (row 1 of $A$ × column 1 of $A^{- 1}$ ): $1 (- 11) + 0 (- 4) + 2 (6) = - 11 + 12 = 1$ ✓
(row 2 of $A$ × column 1): $2 (- 11) + (- 1) (- 4) + 3 (6) = - 22 + 4 + 18 = 0$ ✓

Answer

A^{- 1} = - 11 - 4 6 20 - 1 21 - 1

Problem 5Exam level

If $A = [34 - 2 - 2]$ , verify $A (adj A) = ∣ A ∣ I$ .

Show solution

$∣ A ∣ = 3 (- 2) - (- 2) (4) = - 6 + 8 = 2$

$adj A = [- 2 - 4 23]$

$A (adj A) = [34 - 2 - 2] [- 2 - 4 23] = [- 6 + 8 - 8 + 8 6 - 6 8 - 6] = [2002] = 2 I = ∣ A ∣ I ✓$

Multiplying on the other side gives the same:

$(adj A) A = [- 2 - 4 23] [34 - 2 - 2] = [- 6 + 8 - 12 + 12 4 - 4 8 - 6] = 2 I ✓$

AnswerBoth products equal

2 I = ∣ A ∣ I

Problem 6Challenge

$A$ is a non-singular matrix of order $3$ with $∣ A ∣ = 4$ . Find $∣ adj A ∣$ , $∣ adj (adj A) ∣$ and $∣ A^{- 1} ∣$ .

Show solution

Here $n = 3$ .

$∣ adj A ∣$ . Take determinants of the identity $A (adj A) = ∣ A ∣ I$ :

$∣ A ∣ \cdot ∣ adj A ∣ = ∣ A ∣ I = ∣ A ∣^{n} = ∣ A ∣^{3}$

$\Rightarrow ∣ adj A ∣ = ∣ A ∣^{n - 1} = 4^{2} = 16$

$∣ adj (adj A) ∣$ . Apply the same rule to the matrix $adj A$ , whose determinant is $16$ :

$∣ adj (adj A) ∣ = ∣ adj A ∣^{n - 1} = 1 6^{2} = 256$

(Equivalently $∣ A ∣^{(n - 1)^{2}} = 4^{4} = 256$ .)

$∣ A^{- 1} ∣$ . From $A A^{- 1} = I$ , taking determinants gives $∣ A ∣ ∣ A^{- 1} ∣ = 1$ :

$∣ A^{- 1} ∣ = \frac{1}{4}$

Answer

∣ adj A ∣ = 16

∣ adj (adj A) ∣ = 256

∣ A^{- 1} ∣ = 1/4

Problem 7Challenge

If $A$ and $B$ are non-singular matrices of the same order, prove $(A B)^{- 1} = B^{- 1} A^{- 1}$ .

Show solution

To prove $X$ is the inverse of $A B$ , it is enough to show $(A B) X = I$ and $X (A B) = I$ . Take $X = B^{- 1} A^{- 1}$ .

First product, using associativity throughout:

$(A B) (B^{- 1} A^{- 1}) = A (B B^{- 1}) A^{- 1} = A I A^{- 1} = A A^{- 1} = I$

Second product:

$(B^{- 1} A^{- 1}) (A B) = B^{- 1} (A^{- 1} A) B = B^{- 1} I B = B^{- 1} B = I$

Both products give $I$ , and the inverse is unique, so

$(A B)^{- 1} = B^{- 1} A^{- 1} ■$

Why the order reverses. $B^{- 1}$ must be adjacent to $B$ for the cancellation to happen. Writing $A^{- 1} B^{- 1}$ would leave $A B A^{- 1} B^{- 1}$ , which does not simplify unless $A$ and $B$ happen to commute.

Answer

(A B) (B^{- 1} A^{- 1}) = (B^{- 1} A^{- 1}) (A B) = I

, so

(A B)^{- 1} = B^{- 1} A^{- 1}