JEE Mains Maths · Matrices

Adjoint, Inverse & Determinant Identities

The identity toolkit that turns adjoint, inverse and determinant questions into one-line exponent arithmetic — det(AB)=det A·det B, |adj A|=|A|ⁿ⁻¹, adj(adj A)=|A|ⁿ⁻² A, and A⁻¹=adj A/|A|.

All Matrices notes JEE Mains Maths strategy

Why this matters

Twenty-five PYQs, and JEE Mains repeats this cluster almost every session. Very few of them ask you to actually compute an adjoint or an inverse entry-by-entry — the paper rewards students who KNOW the identities cold and finish in three lines of exponent arithmetic. The heaviest hitter is the adjoint-of-adjoint family: for a 3×3 matrix, adj(adj A)=|A|·A and |adj(adj A)|=|A|⁴, and getting the order n and the power right is the whole question. The rest split between the Cayley–Hamilton route to a 2×2 inverse (A⁻¹=αA+βI), invertibility conditions (det≠0), and a few counting/special-inverse twists. Seven concepts cover every one.

Concept 1 of 7

Determinant of products, transposes and scalar multiples

Intuition

The determinant is multiplicative: it splits across products, ignores transpose, and pulls a scalar out as

k^n

. These three facts collapse a fearsome-looking product like

A^TC^2A

into a product of numbers.

Definition

For $n\times n$ matrices:

Product: $\det(AB) = \det A\,\det B$
Transpose: $\det(A^T) = \det A$
Scalar: $\det(kA) = k^n\det A$ (the order $n$ is the exponent)
Power: $\det(A^m) = (\det A)^m$ , and $\det(A^{-1}) = 1/\det A$

Determinant identities

\det(AB)=\det A\,\det B,\quad \det(A^T)=\det A,\quad \det(kA)=k^{\,n}\det A

Worked example

Let

A,B

3\times3

matrices with

|A| = 2

and

|B| = 5

. Find

|2A^TB|

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1MatricesMODERATE

A =\begin{bmatrix} \sqrt{2} & 1 \\ - 1 & \sqrt{2} \end{bmatrix},B =\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix},C =ABA^{T}

and

X

=A^{T}C^{2}\text{ }A

, then

detX

is equal to:

[Q63 · 1 February 2024 · 2024]

$\det(kA) = k^{\,n}\det A$ , NOT $k\det A$

Every scalar multiple of an

n\times n

matrix multiplies the determinant by

k^n

— one factor of

k

per row. For a

3\times3

matrix,

|2A| = 8|A|

, not

2|A|

. Forgetting the power

n

is the single most common determinant slip.

Drill 1 more on determinant of products, transposes and scalar multiples

Concept 2 of 7

Adjoint identities: A·adj A = |A|I and |adj A| = |A|ⁿ⁻¹

Intuition

You almost never compute an adjoint on JEE Mains. Instead you use its two defining facts: multiplying

A

by its adjoint gives

|A|I

, and the determinant of the adjoint is a single power of

|A|

. Those two identities close most 'adjoint' questions in one line.

Definition

For an $n\times n$ matrix:

Defining identity: $A(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|\,I_n$
Determinant: $|\operatorname{adj}A| = |A|^{\,n-1}$
Scalar: $\operatorname{adj}(kA) = k^{\,n-1}\operatorname{adj}A$
2×2 shortcut: $\operatorname{adj}\begin{pmatrix}a&b\\c&d\end{pmatrix} = \begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ (swap the diagonal, negate the off-diagonal)

Adjoint identities

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

Worked example

For a

3\times3

matrix

A

with

|A| = 4

, find

|\operatorname{adj}A|

, and state

A(\operatorname{adj}A)

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2MatricesMODERATE

Let

\mathbf{S = \{}\sqrt{n}:1 \leq n \leq 50

and

n

is odd

\}

Let

a \in S

and

A =\begin{bmatrix} 1 & 0 & a \\ - 1 & 1 & 0 \\ - a & 0 & 1 \end{bmatrix}

\sum_{a \in S} det(adjA) = 100\lambda

, then

\lambda

is equal to

[Q69 · 14 June 2022 · 2022]

$|\operatorname{adj}A| = |A|^{\,n-1}$ — the exponent is $n-1$ , NOT $n$

For a

3\times3

matrix,

|\operatorname{adj}A| = |A|^2

, not

|A|^3

and not

|A|

. Students who write

|A|^3

(over-counting) or

|A|

(dropping the power) walk into the two standard distractors.

$\operatorname{adj}(kA) = k^{\,n-1}\operatorname{adj}A$ — note $n-1$ , not $n$

The adjoint scales by

k^{n-1}

, one power LESS than the determinant's

k^n

. For a

3\times3

\operatorname{adj}(2A) = 4\operatorname{adj}A

, not

8\operatorname{adj}A

. Mixing up which one loses the power of

n

is the trap.

Drill 4 more on adjoint identities: a·adj a = |a|i and |adj a| = |a|ⁿ⁻¹

Concept 3 of 7

Adjoint of the adjoint

Intuition

This is the most-tested identity in the whole subtopic. The double adjoint gives back

A

scaled by a power of its determinant, and its determinant is a single high power of

|A|

. For a

3\times3

matrix the numbers are clean:

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

and

|\operatorname{adj}(\operatorname{adj}A)| = |A|^4

Definition

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

Double adjoint: $\operatorname{adj}(\operatorname{adj}A) = |A|^{\,n-2}A$
Its determinant: $|\operatorname{adj}(\operatorname{adj}A)| = |A|^{\,(n-1)^2}$

For $n = 3$ : $\operatorname{adj}(\operatorname{adj}A) = |A|\,A$ and $|\operatorname{adj}(\operatorname{adj}A)| = |A|^{4}$ . For $n = 2$ : $\operatorname{adj}(\operatorname{adj}A) = A$ .

Double adjoint

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

Worked example

For a

3\times3

matrix

A

with

|A| = 3

, find

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

and

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

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3MatricesMODERATE

For a

3 \times 3

matrix M , let trace

(M)

denote the sum of all the diagonal elements of M . Let A be a

3 \times 3

matrix such that

|A| =\frac{1}{2}

and trace

(A) = 3

. If

B = adj(adj(2A))

, then the value of

|B| + trace(B)

equals:

[Q129 · 22 Jan 2025 · 2025]

$\operatorname{adj}(\operatorname{adj}A) = |A|^{\,n-2}A$ : for $3\times3$ it's $|A|\cdot A$ , for $2\times2$ it's just $A$

The power on

|A|

n-2

. For

n = 3

that is

|A|^1

, so

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

; for

n = 2

it is

|A|^0 = 1

, so

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

. Writing

|A|A

for a

2\times2

is a classic over-application.

$|\operatorname{adj}(\operatorname{adj}A)| = |A|^{4}$ for $3\times3$ , NOT $|A|^2$

The determinant of the double adjoint carries the exponent

(n-1)^2

. For a

3\times3

that is

(3-1)^2 = 4

, giving

|A|^4

. It is easy to stop at

|A|^2

(the single-adjoint power) or at

|A|^{n-2}=|A|

(the matrix power) — both are wrong for the determinant.

Drill 4 more on adjoint of the adjoint

Concept 4 of 7

Inverse from the adjoint and PQ = kI

Intuition

The inverse is the adjoint divided by the determinant. Two common disguises:

AB = I

means

B = A^{-1}

, and

PQ = kI

means

Q = kP^{-1}

— from which the determinant

|Q| = k^n/|P|

drops out immediately.

Definition

$A^{-1} = \dfrac{\operatorname{adj}A}{|A|}$ , defined iff $|A| \neq 0$ . Consequences used in PYQs:

From a product = I: $AB = I \Rightarrow B = A^{-1}$ , so $|B| = 1/|A|$
**From $PQ = kI_n$ :** $Q = kP^{-1}$ , so $|Q| = \dfrac{k^{\,n}}{|P|}$
Adjoint from inverse: $\operatorname{adj}A = |A|\,A^{-1}$ (rearranging the defining identity)

Inverse and PQ = kI

A^{-1} = \frac{\operatorname{adj}A}{|A|}\ (|A|\neq0), \qquad PQ = kI_n \Rightarrow Q = kP^{-1},\ \ |Q| = \frac{k^{\,n}}{|P|}

Worked example

Let

P

be a

3\times3

matrix with

|P| = 4

, and let

Q

satisfy

PQ = 2I_3

. Find

|Q|

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4MatricesMODERATE

Let

P=\begin{bmatrix} 3 & - 1 & - 2 \\ 2 & 0 & \alpha \\ 3 & - 5 & 0 \end{bmatrix}

, where

\alpha\in R

. Suppose

Q=\left\lbrack q_{ij} \right\rbrack

is a matrix satisfying

PQ=kI_{3}

for some non-zero

k\in R

. If

q_{23}= -\frac{k}{8}

and

|Q| =\frac{k^{2}}{2}

, then

\alpha^{2}+k^{2}

is equal to

[Q81 · Paper 1 · 2021]

$PQ = kI_n \Rightarrow |Q| = k^{\,n}/|P|$ — the $k^n$ , not $k$

From

Q = kP^{-1}

, taking determinants pulls the scalar out as

k^n

(order

n

), so

|Q| = k^n|P^{-1}| = k^n/|P|

. For a

3\times3

, that is

k^3/|P|

. Writing

k/|P|

forgets the

k^n

rule and is the intended distractor.

Drill 1 more on inverse from the adjoint and pq = ki

Concept 5 of 7

Inverse via Cayley–Hamilton (A⁻¹ = αA + βI)

Intuition

Every

2\times2

matrix satisfies its own characteristic equation

A^2 - (\operatorname{tr}A)A + (\det A)I = O

. Multiply through by

A^{-1}

and the inverse becomes a linear combination of

A

and

I

— no adjoint needed. The same trick handles a nilpotent

X

(

X^3 = O

) via the finite series for

(\alpha I + \beta X + \gamma X^2)^{-1}

Definition

2×2 Cayley–Hamilton: $A^2 - (\operatorname{tr}A)A + (\det A)I = O$ . Multiplying by $A^{-1}$ :

A^{-1} = \frac{(\operatorname{tr}A)\,I - A}{\det A}

so in the form

A^{-1} = \alpha A + \beta I

\alpha = -\dfrac{1}{\det A}

and

\beta = \dfrac{\operatorname{tr}A}{\det A}

. Nilpotent case: if

X^3 = O

then

Y = \alpha I + \beta X + \gamma X^2

has

Y^{-1} = \tfrac{1}{\alpha}\big(I - \tfrac{\beta}{\alpha}X + (\tfrac{\beta^2}{\alpha^2}-\tfrac{\gamma}{\alpha})X^2\big)

— a terminating series because powers

\ge 3

vanish.

2×2 inverse from Cayley–Hamilton

A^{2} - (\operatorname{tr}A)A + (\det A)I = O \ \Rightarrow\ A^{-1} = \frac{(\operatorname{tr}A)I - A}{\det A}

Worked example

For

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

, write

A^{-1} = \alpha A + \beta I

and find

\alpha,\beta

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5MatricesMODERATE

A=\begin{bmatrix} 1 & 2 \\ - 1 & 4 \end{bmatrix}A^{- 1}=\alpha I+\beta A

, find

5(\alpha-\beta) =

[Q56 · 2021 Compilation Paper 16 · 2021]

Divide by $\det A$ — Cayley–Hamilton gives $A^{-1}$ , not $\operatorname{adj}A$

After multiplying

A^2 - (\operatorname{tr}A)A + (\det A)I = O

A^{-1}

, you get

(\det A)A^{-1} = (\operatorname{tr}A)I - A

. You MUST divide by

\det A

to isolate

A^{-1}

; stopping at

(\operatorname{tr}A)I - A

gives

\operatorname{adj}A

(unnormalised), off by a factor of

\det A

Drill 3 more on inverse via cayley–hamilton (a⁻¹ = αa + βi)

Concept 6 of 7

When is a matrix invertible? (det ≠ 0)

Intuition

A square matrix is invertible exactly when its determinant is non-zero. That single test answers 'for which

t

…', 'does the system have a unique solution', and 'is

A - \lambda I

invertible'. And once

AB = I

for square

A,B

, the inverse is two-sided:

BA = I

as well.

Definition

An $n\times n$ matrix $A$ is invertible (non-singular) $\iff |A| \neq 0$ .

Linear system: $Ax = b$ has a unique solution $\iff |A| \neq 0$ .
Two-sided inverse: for square $A,B$ , $AB = I \Rightarrow BA = I$ (a one-sided inverse is automatically two-sided in finite dimension).
Deducing a determinant from an invertible factor: if $MN = O$ and $N$ is invertible, then $M = O$ (multiply by $N^{-1}$ ); hence $|M| = 0$ .

Invertibility test

A^{-1}\text{ exists} \iff |A| \neq 0 \qquad\text{and}\qquad AB = I \Rightarrow BA = I \ \text{(square matrices)}

Worked example

For which real

t

M = \begin{pmatrix}t & 1\\4 & t\end{pmatrix}

invertible?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 6MatricesMODERATE

Let

A

and

B

be two

3 \times 3

real matrices such that

\left( A^{2}-B^{2} \right)

is invertible matrix. If

A^{5}=B^{5}

and

A^{3}B^{2}=A^{2}B^{3}

, then the value of the determinant of the matrix

A^{3}+B^{3}

is equal to:

[Q79 · Paper 17 · 2021]

$MN = O$ with $N$ invertible forces $M = O$ (so $|M| = 0$ )

If a product is the zero matrix and one factor is invertible, the OTHER factor must be zero — multiply

MN = O

on the right by

N^{-1}

. Questions like '

A^2 - B^2

invertible, find

\det(A^3 + B^3)

' collapse to

A^3 + B^3 = O

, so the determinant is

0

. Don't try to compute entries.

Drill 3 more on when is a matrix invertible? (det ≠ 0)

Concept 7 of 7

Special inverses, counting, and the reversal law

Intuition

A grab-bag of inverse twists: the reversal law flips the order of a product's inverse, involutory matrices satisfy

A = A^{-1}

(i.e.

A^2 = I

) so counting them means counting

A^2 = I

, and for a

2\times2

the determinant of

A + I

has a neat trace formula.

Definition

Reversal law: $(AB)^{-1} = B^{-1}A^{-1}$ and $\operatorname{adj}(AB) = \operatorname{adj}B\,\operatorname{adj}A$ ; with $\operatorname{adj}M = |M|M^{-1}$ , expressions like $A(\operatorname{adj}A^{-1} + \operatorname{adj}B^{-1})^{-1}B$ simplify to a multiple of $\operatorname{adj}A + \operatorname{adj}B$ .
Involutory / self-inverse: $A = A^{-1} \iff A^2 = I$ ; counting such matrices means counting solutions of $A^2 = I$ .
2×2 determinant shift: $\det(A + I) = \det A + \operatorname{tr}A + 1$ (and $\det(\operatorname{adj}A + I) = \det A + \operatorname{tr}A + 1$ too, since $\operatorname{tr}(\operatorname{adj}A) = \operatorname{tr}A$ ).

Reversal law and self-inverse

(AB)^{-1} = B^{-1}A^{-1}, \qquad A = A^{-1} \iff A^{2} = I, \qquad \det(A + I) = \det A + \operatorname{tr}A + 1\ (2\times2)

Worked example

Let

A

be a

2\times2

matrix with

\det A = -1

. Simplify

\det\big((A + I)(\operatorname{adj}A + I)\big)

in terms of

\operatorname{tr}A

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 7MatricesMODERATE

The number of matrices

A =\begin{bmatrix} a & b \\ c & d \end{bmatrix}

, where a, b,c,d

\mathbf{\in \{ - 1,0,1,2,3,\ldots\ldots,10\}}

, such that

A =A^{- 1}

, is

[Q180 · 26 July 2022 · 2022]

$A = A^{-1}$ means $A^2 = I$ , NOT $A = I$

Self-inverse (involutory) matrices are far more than just the identity — any matrix with

A^2 = I

qualifies, including

-I

, reflections, and many trace-zero

2\times2

matrices. A counting question '

A = A^{-1}

' asks for ALL solutions of

A^2 = I

, which can number in the dozens.

$\det(A + I) = \det A + \operatorname{tr}A + 1$ for $2\times2$ , NOT $\det A + 1$

Adding

I

to a

2\times2

shifts the determinant by

\operatorname{tr}A + 1

, not just

1

— the trace term is easy to drop. This is the crux of the

(A + I)(\operatorname{adj}A + I)

family of questions.

Drill 2 more on special inverses, counting, and the reversal law

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 (7)

Determinant of products, transposes and scalar multiples
Determinant identities
$\det(AB)=\det A\,\det B,\quad \det(A^T)=\det A,\quad \det(kA)=k^{\,n}\det A$
Adjoint identities: A·adj A = |A|I and |adj A| = |A|ⁿ⁻¹
Adjoint identities
$A(\operatorname{adj}A) = |A|\,I_n, \qquad |\operatorname{adj}A| = |A|^{\,n-1}, \qquad \operatorname{adj}(kA) = k^{\,n-1}\operatorname{adj}A$
Adjoint of the adjoint
Double adjoint
$\operatorname{adj}(\operatorname{adj}A) = |A|^{\,n-2}A, \qquad |\operatorname{adj}(\operatorname{adj}A)| = |A|^{\,(n-1)^2}$
Inverse from the adjoint and PQ = kI
Inverse and PQ = kI
$A^{-1} = \frac{\operatorname{adj}A}{|A|}\ (|A|\neq0), \qquad PQ = kI_n \Rightarrow Q = kP^{-1},\ \ |Q| = \frac{k^{\,n}}{|P|}$
Inverse via Cayley–Hamilton (A⁻¹ = αA + βI)
2×2 inverse from Cayley–Hamilton
$A^{2} - (\operatorname{tr}A)A + (\det A)I = O \ \Rightarrow\ A^{-1} = \frac{(\operatorname{tr}A)I - A}{\det A}$
When is a matrix invertible? (det ≠ 0)
Invertibility test
$A^{-1}\text{ exists} \iff |A| \neq 0 \qquad\text{and}\qquad AB = I \Rightarrow BA = I \ \text{(square matrices)}$
Special inverses, counting, and the reversal law
Reversal law and self-inverse
$(AB)^{-1} = B^{-1}A^{-1}, \qquad A = A^{-1} \iff A^{2} = I, \qquad \det(A + I) = \det A + \operatorname{tr}A + 1\ (2\times2)$

Watch out for (10)

$\det(kA) = k^{\,n}\det A$ , NOT $k\det A$
→ Determinant of products, transposes and scalar multiples
$|\operatorname{adj}A| = |A|^{\,n-1}$ — the exponent is $n-1$ , NOT $n$
→ Adjoint identities: A·adj A = |A|I and |adj A| = |A|ⁿ⁻¹
$\operatorname{adj}(kA) = k^{\,n-1}\operatorname{adj}A$ — note $n-1$ , not $n$
→ Adjoint identities: A·adj A = |A|I and |adj A| = |A|ⁿ⁻¹
$\operatorname{adj}(\operatorname{adj}A) = |A|^{\,n-2}A$ : for $3\times3$ it's $|A|\cdot A$ , for $2\times2$ it's just $A$
→ Adjoint of the adjoint
$|\operatorname{adj}(\operatorname{adj}A)| = |A|^{4}$ for $3\times3$ , NOT $|A|^2$
→ Adjoint of the adjoint
$PQ = kI_n \Rightarrow |Q| = k^{\,n}/|P|$ — the $k^n$ , not $k$
→ Inverse from the adjoint and PQ = kI
Divide by $\det A$ — Cayley–Hamilton gives $A^{-1}$ , not $\operatorname{adj}A$
→ Inverse via Cayley–Hamilton (A⁻¹ = αA + βI)
$MN = O$ with $N$ invertible forces $M = O$ (so $|M| = 0$ )
→ When is a matrix invertible? (det ≠ 0)
$A = A^{-1}$ means $A^2 = I$ , NOT $A = I$
→ Special inverses, counting, and the reversal law
$\det(A + I) = \det A + \operatorname{tr}A + 1$ for $2\times2$ , NOT $\det A + 1$
→ Special inverses, counting, and the reversal law

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