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.

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

Try it yourself

What is the sign of the cofactor

C_{23}

, and what is

a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Cofactor sign at position (1,2)?
2.
$\sum_j a_{2j}C_{2j} = ?$
3.
$\sum_j a_{1j}C_{2j} = ?$
4.
Minor $M_{ij}$ is found by?

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

Try it yourself

What is

A(\operatorname{adj}A)

for a

3\times3

matrix with

|A| = 5

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\operatorname{adj}A = ?$ (in terms of cofactors)
2.
$A(\operatorname{adj}A) = ?$
3.
$\operatorname{adj}\begin{pmatrix}1&2\\3&4\end{pmatrix}$ ?
4.
Adjoint of identity $I_3$ ?

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 5 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: $|\operatorname{adj}A| = |A|^{n-1}$ ; $\operatorname{adj}(\operatorname{adj}A) = |A|^{n-2}A$ ; $\operatorname{adj}(kA) = k^{n-1}\operatorname{adj}A$ ; $\operatorname{adj}(AB) = \operatorname{adj}B\,\operatorname{adj}A$ (reversal).

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 · 4 quick reps

Try it yourself

For a

3\times3

matrix

A

with

|A| = 4

, find

|2\,\operatorname{adj}(3A)|

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$|\operatorname{adj}A|$ for 3×3 with $|A| = 3$ ?
2.
$\operatorname{adj}(\operatorname{adj}A)$ for 3×3?
3.
$\operatorname{adj}(kA)$ for n×n?
4.
$\operatorname{adj}(AB) = ?$

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]

Drill 6 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

Try it yourself

Find

A^{-1}

for

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

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$A^{-1} = ?$ in terms of adjoint
2.
Inverse exists iff?
3.
Inverse of $\operatorname{diag}(2,5)$ ?
4.
$|A^{-1}|$ in terms of $|A|$ ?

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

$(AB)^{-1} = B^{-1}A^{-1}$ (reversal); $(A^{-1})^{-1} = A$ ; $\det(A^{-1}) = 1/\det A$ ; $(A^T)^{-1} = (A^{-1})^T$ ; $(kA)^{-1} = \tfrac1k A^{-1}$ .

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

Try it yourself

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

correct? And what is

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

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$(AB)^{-1} = ?$
2.
$(A^{-1})^{-1} = ?$
3.
$\det(A^{-1}) = ?$
4.
$(A^T)^{-1} = ?$

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]

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.

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

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Inverse of an orthogonal matrix?
2.
Inverse of $\operatorname{diag}(3,4)$ ?
3.
Inverse of a rotation $R(\theta)$ ?
4.
Inverse of an involutory matrix ( $A^2=I$ )?

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 (5)

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

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Matrices & DeterminantsHARD

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

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

[Q30 · Sep · 2022]

Example 2Matrices & DeterminantsHARD

Let

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

What is

A^{-1}

equal to?

[Q50 · Sep · 2024]

Example 3Matrices & DeterminantsHARD

Let

A

be a matrix of order

3\times3

and

|A|=4

. If

|2\,\text{adj}(3A)|=2^{\alpha}3^{\beta}

, then what is the value of

(\alpha+\beta)

[Q4 · Apr · 2023]

Example 4Matrices & DeterminantsMODERATE

A = \begin{pmatrix}2\sin\theta & \cos\theta & 0\\-2\cos\theta & \sin\theta & 0\\-1 & 1 & 1\end{pmatrix}

, then what is

A(\text{adj}A)

equal to?

[Q5 · Apr · 2022]

Example 5Matrices & DeterminantsMODERATE

Consider the following statements in respect of two nonsingular matrices

A

and

B

of the same order

n

: A.

\text{adj}(AB)=(\text{adj }A)(\text{adj }B)

\text{adj}(AB)=\text{adj}(BA)

(AB)\,\text{adj}(AB)-|AB|I_n

is a null matrix of order

n

How many of the above statements are correct?

[Q11 · Apr · 2024]

Drill every past-year question on this subtopic

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

Drill the 28 questions