NDA Maths · Matrices & Determinants

Cofactors, Adjoint & Inverse

Minors and cofactors build the adjoint; the adjoint over the determinant gives the inverse — and the powers of the determinant (|adj A| = |A|ⁿ⁻¹) answer most of the rest.

All Matrices & Determinants notes NDA Maths strategy

Why this matters

Twenty-eight PYQs spanning EASY to HARD. This is the machinery linking determinants to inverses. Questions test cofactor expansion (and the alien-cofactor = 0 trap), adjoint computation, the adjoint power formulas, the inverse via adjoint, the reversal law (AB)⁻¹ = B⁻¹A⁻¹, and the easy inverses of diagonal/orthogonal matrices. Six concepts cover it.

Test yourself — a quick 14-question Matrices & Determinants recall quiz, instant score.

Concept 1 of 6

Minors, cofactors, and expansion

Intuition

The minor

M_{ij}

is the determinant left after deleting row

i

and column

j

; the cofactor attaches the sign

(-1)^{i+j}

. Expanding a row against its OWN cofactors gives the determinant; against ANOTHER row's cofactors (alien cofactors) gives 0.

Definition

Cofactor $C_{ij} = (-1)^{i+j}M_{ij}$ . Own-row expansion: $\sum_j a_{ij}C_{ij} = \det A$ . Alien cofactors: $\sum_j a_{ij}C_{kj} = 0$ when $k \neq i$ (a row times another row's cofactors). The sign pattern is the checkerboard $(-1)^{i+j}$ .

Cofactor expansion vs alien cofactors

\sum_j a_{ij}C_{ij} = \det A, \qquad \sum_j a_{ij}C_{kj} = 0\ (k \neq i)

Worked example

For

A = \begin{pmatrix}2&1&3\\0&5&4\\1&2&1\end{pmatrix}

, find the cofactor

C_{31}

Minor $M_{31}$ : delete row 3 and column 1, take the determinant: $\begin{vmatrix}1&3\\5&4\end{vmatrix} = 4 - 15 = -11$ .
Cofactor $C_{31} = (-1)^{3+1} M_{31} = (+1)(-11)$ .
$= -11$ .

Answer:

C_{31} = -11

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Matrices & DeterminantsEASY

Consider the following for the items that follow: Let

\Delta

be the determinant of a matrix A, where

A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}

and

C_{11}, C_{12}, C_{13}

be the cofactors of

a_{11}, a_{12}, a_{13}

respectively.

What is the value of

a_{21}C_{11} + a_{22}C_{12} + a_{23}C_{13}

[Q29 · Sep · 2022]

Drill 3 more on minors, cofactors, and expansion

Concept 2 of 6

The adjoint (adjugate)

Intuition

The adjoint is the TRANSPOSE of the cofactor matrix. Its defining property —

A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|\,I

— is the bridge to the inverse. For

2\times2

there's a one-line shortcut: swap the diagonal, negate the off-diagonal.

Definition

$\operatorname{adj}A = [C_{ij}]^T$ (transpose of the cofactor matrix). Defining identity: $A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|\,I_n$ . 2×2 shortcut: $\operatorname{adj}\begin{pmatrix}a&b\\c&d\end{pmatrix} = \begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ .

Adjoint identity

A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|\,I_n

Worked example

Find the adjoint of

A = \begin{pmatrix}3 & 1\\2 & 4\end{pmatrix}

Use the 2×2 shortcut: swap diagonal entries (3 and 4), negate the off-diagonal (1 and 2).
$\operatorname{adj}A = \begin{pmatrix}4 & -1\\-2 & 3\end{pmatrix}$ .
Check: $A(\operatorname{adj}A) = (12 - 2)I = 10 I = |A|I$ . ✓

Answer:

\operatorname{adj}A = \begin{pmatrix}4 & -1\\-2 & 3\end{pmatrix}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Matrices & DeterminantsHARD

The adjoint of the matrix

A = \begin{pmatrix}1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 3 & 1\end{pmatrix}

[Q97 · Sep · 2017]

Drill 4 more on the adjoint (adjugate)

Concept 3 of 6

Adjoint power formulas

Intuition

Almost every 'adjoint of adjoint' or '|adj A|' question is a single power formula. The keys:

|\operatorname{adj}A| = |A|^{n-1}

and

\operatorname{adj}(\operatorname{adj}A) = |A|^{n-2}A

for an

n\times n

matrix.

Definition

For an $n\times n$ non-singular matrix:

Determinant: $|\operatorname{adj}A| = |A|^{n-1}$
Double adjoint: $\operatorname{adj}(\operatorname{adj}A) = |A|^{n-2}A$
Scalar: $\operatorname{adj}(kA) = k^{n-1}\operatorname{adj}A$
Reversal: $\operatorname{adj}(AB) = \operatorname{adj}B\,\operatorname{adj}A$
Transpose: $\operatorname{adj}(A^T) = (\operatorname{adj}A)^T$ (adjoint commutes with transpose, so $\operatorname{adj}A^T - (\operatorname{adj}A)^T = O$ )

Adjoint of an n×n matrix

|\operatorname{adj}A| = |A|^{\,n-1}, \qquad \operatorname{adj}(\operatorname{adj}A) = |A|^{\,n-2}A

Worked example

For a

3\times3

matrix

A

with

|A| = 3

, find

|\operatorname{adj}(\operatorname{adj}A)|

For a $3\times3$ matrix, $\operatorname{adj}(\operatorname{adj}A) = |A|^{n-2}A = |A|^{1}A = 3A$ .
So $|\operatorname{adj}(\operatorname{adj}A)| = |3A| = 3^3\,|A| = 27 \cdot 3$ .
$= 81$ .

Answer:

81

(

= |A|^{(n-1)^2} = 3^4

Practice this conceptself-check · 5 quick reps

From the bank · past-year question

Example 3Matrices & DeterminantsMODERATE

M=\begin{pmatrix}2&0&0\\0&2&0\\0&0&2\end{pmatrix}

, then what is the value of

|M|\cdot|\text{adj}M|

[Q19 · Apr · 2026]

$|\operatorname{adj}A| = |A|^{\,n-1}$ , NOT $|A|$ or $|A|^n$

The determinant of the adjoint carries the exponent

n-1

, where

n

is the ORDER. For a

3\times3

matrix it is

|A|^2

— students who answer

|A|

(forgetting the power) or

|A|^3

(over-counting) walk into the two standard distractors.

$\operatorname{adj}(AB) = \operatorname{adj}B\,\operatorname{adj}A$ — the order REVERSES

Like transpose and inverse, the adjoint of a product flips the factors:

\operatorname{adj}(AB) = \operatorname{adj}B\,\operatorname{adj}A

, not

\operatorname{adj}A\,\operatorname{adj}B

. Keeping the original order is the trap.

Drill 7 more on adjoint power formulas

Concept 4 of 6

Inverse via the adjoint

Intuition

Divide the adjoint by the determinant and you have the inverse:

A^{-1} = \frac{1}{|A|}\operatorname{adj}A

. It exists exactly when

|A| \neq 0

(non-singular).

Definition

$A^{-1} = \dfrac{1}{|A|}\operatorname{adj}A$ , defined iff $|A| \neq 0$ . For $2\times2$ : $A^{-1} = \dfrac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ .

Inverse formula

A^{-1} = \frac{1}{|A|}\operatorname{adj}A \quad (|A| \neq 0)

Worked example

Find the inverse of

A = \begin{pmatrix}3&5\\1&2\end{pmatrix}

using the adjoint.

$|A| = 3\cdot2 - 5\cdot1 = 1$ .
$\operatorname{adj}A = \begin{pmatrix}2&-5\\-1&3\end{pmatrix}$ (swap the diagonal, negate the off-diagonal).
$A^{-1} = \dfrac{1}{|A|}\operatorname{adj}A = \begin{pmatrix}2&-5\\-1&3\end{pmatrix}$ .

Answer:

A^{-1} = \begin{pmatrix}2&-5\\-1&3\end{pmatrix}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Matrices & DeterminantsEASY

a, b, c

are non-zero real numbers, then the inverse of the matrix

A = \begin{pmatrix}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{pmatrix}

is equal to

[Q19 · Sep · 2017]

Drill 3 more on inverse via the adjoint

Concept 5 of 6

Inverse properties and the reversal law

Intuition

The inverse of a product reverses the order —

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

— exactly like transpose and adjoint. Other staples: inverting twice returns

A

, and the determinant of the inverse is the reciprocal.

Definition

Key inverse identities:

Reversal: $(AB)^{-1} = B^{-1}A^{-1}$ (order flips)
Double inverse: $(A^{-1})^{-1} = A$
Determinant: $\det(A^{-1}) = 1/\det A$
Transpose: $(A^T)^{-1} = (A^{-1})^T$
Scalar: $(kA)^{-1} = \tfrac1k A^{-1}$

Reversal law for inverses

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

Worked example

A

and

B

are invertible of the same order, what is

(AB)^{-1}

The inverse of a product reverses the order of factors.
Check: $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = I$ . ✓

Answer:

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

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Matrices & DeterminantsEASY

If A and B are two invertible square matrices of same order, then what is

(AB)^{-1}

equal to?

[Q18 · Sep · 2018]

$(AB)^{-1} = B^{-1}A^{-1}$ , NOT $A^{-1}B^{-1}$

The inverse of a product reverses the order of factors — the single most-missed inverse fact. It must, so that

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

. Writing

A^{-1}B^{-1}

is the trap.

$(A^T)^{-1} = (A^{-1})^T$ — don't drop the transpose

Transpose and inverse commute:

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

. The slip is to compute

A^{-1}

and forget to transpose it (or vice-versa), reporting plain

A^{-1}

as the answer.

Drill 5 more on inverse properties and the reversal law

Concept 6 of 6

Inverses of special matrices

Intuition

Some inverses need no adjoint at all: a diagonal matrix inverts entrywise, and an orthogonal (or rotation) matrix has

A^{-1} = A^T

. Recognise the type and write the inverse down.

Definition

Diagonal: $\operatorname{diag}(d_i)^{-1} = \operatorname{diag}(1/d_i)$ . Orthogonal / rotation: $A^{-1} = A^T$ (and for a rotation $R(\theta)^{-1} = R(-\theta)$ ). Involutory: $A^{-1} = A$ . Use these instead of the adjoint route when the type is clear.

Inverse properties of special matrices

(A^{-1})^{-1} = A \qquad (A^T)^{-1} = (A^{-1})^T \qquad (kA)^{-1} = \frac{1}{k}A^{-1} \qquad (A^n)^{-1} = (A^{-1})^n

Worked example

Find the inverse of

A = \begin{pmatrix}0&1\\1&0\end{pmatrix}

Compute $A^2 = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I$ .
$A^2 = I$ means $A$ is involutory, so $A^{-1} = A$ .

Answer:

A^{-1} = A = \begin{pmatrix}0&1\\1&0\end{pmatrix}

Practice this concept4 quick reps

From the bank · past-year question

Example 6Matrices & DeterminantsMODERATE

What is the inverse of the matrix

A = \begin{pmatrix}\cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1\end{pmatrix}

[Q24 · Apr · 2018]

Drill 1 more on inverses of special matrices

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (6)

Minors, cofactors, and expansion
Cofactor expansion vs alien cofactors
$\sum_j a_{ij}C_{ij} = \det A, \qquad \sum_j a_{ij}C_{kj} = 0\ (k \neq i)$
The adjoint (adjugate)
Adjoint identity
$A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|\,I_n$
Adjoint power formulas
Adjoint of an n×n matrix
$|\operatorname{adj}A| = |A|^{\,n-1}, \qquad \operatorname{adj}(\operatorname{adj}A) = |A|^{\,n-2}A$
Inverse via the adjoint
Inverse formula
$A^{-1} = \frac{1}{|A|}\operatorname{adj}A \quad (|A| \neq 0)$
Inverse properties and the reversal law
Reversal law for inverses
$(AB)^{-1} = B^{-1}A^{-1}$
Inverses of special matrices
Inverse properties of special matrices
$(A^{-1})^{-1} = A \qquad (A^T)^{-1} = (A^{-1})^T \qquad (kA)^{-1} = \frac{1}{k}A^{-1} \qquad (A^n)^{-1} = (A^{-1})^n$

Watch out for (4)

$|\operatorname{adj}A| = |A|^{\,n-1}$ , NOT $|A|$ or $|A|^n$
→ Adjoint power formulas
$\operatorname{adj}(AB) = \operatorname{adj}B\,\operatorname{adj}A$ — the order REVERSES
→ Adjoint power formulas
$(AB)^{-1} = B^{-1}A^{-1}$ , NOT $A^{-1}B^{-1}$
→ Inverse properties and the reversal law
$(A^T)^{-1} = (A^{-1})^T$ — don't drop the transpose
→ Inverse properties and the reversal law

Drill every past-year question on this subtopic

28 questions from the bank — paginated, with cart and Word-export support.

Drill the 28 questions