JEE Mains Maths · Formula sheet

Matrices formulas

27 formulas and 33 common traps for JEE Mains Maths Matrices, grouped by subtopic.

Matrix Algebra, Types & Operations

Order, equality, linear combinations, and trace

Trace (sum of the diagonal)

\operatorname{tr}(A)=\sum_{i=1}^{n} a_{ii}

Matrix multiplication: conformability and the row-by-column rule

Row-by-column entry rule

(AB)_{ij}=\sum_{k} a_{ik}\,b_{kj}

Transpose and the reversal law

Transpose rules

(AB)^{T}=B^{T}A^{T}\qquad (A^{T})^{T}=A

Non-commutativity, zero divisors, and commuting matrices

Non-commutative square expansion

(A+B)^{2}=A^{2}+AB+BA+B^{2}

Elementary row operations

The three operations

R_i\leftrightarrow R_j\qquad R_i\to kR_i\ (k\neq0)\qquad R_i\to R_i+kR_j

Counting matrices

Counting bridges

\#=k^{mn}\qquad \operatorname{tr}(A^{T}A)=\sum_{i,j}a_{ij}^{2}

Common traps

Trace is the DIAGONAL sum, not the sum of all entries

\operatorname{tr}(A)

adds only

a_{11},a_{22},\dots,a_{nn}

— the main diagonal. Students who total every entry of the matrix get the wrong number. Only the diagonal counts, and only square matrices have a trace at all.

Matrix multiplication is NOT commutative — $AB\neq BA$

Order matters:

AB

and

BA

are generally different matrices, and one may be undefined while the other exists. Never swap factors inside a product. If a question defines

AB

, compute

AB

— reaching for

BA

is the classic slip.

$(AB)^{T}=B^{T}A^{T}$ — the order REVERSES

The transpose of a product flips the factors:

(AB)^{T}=B^{T}A^{T}

, never

A^{T}B^{T}

. Sum and scalar transposes keep their order, so only the PRODUCT reverses. Writing

A^{T}B^{T}

is the trap answer.

$AB=O$ does NOT force $A=O$ or $B=O$

Unlike numbers, matrices have zero divisors: two nonzero matrices can multiply to

O

(e.g.

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

). You may not 'cancel' a matrix. When

A,B

are nonzero with

AB=O

, the correct conclusion is that both are singular (

\det=0

Don't import $a^{2}-b^{2}=(a+b)(a-b)$ into matrices

Every number identity that secretly uses commutativity can fail for matrices:

(A+B)(A-B)

(A+B)^{2}

, and

(AB)^{2}=A^{2}B^{2}

all break unless

AB=BA

. An option that quietly assumes one of these is almost always the trap.

Powers of a Matrix & the Cayley-Hamilton Theorem

Learn this subtopic in the notes

Cyclic and pattern powers

Cyclic reduction

A^d = I \ \Rightarrow\ A^n = A^{\,n \bmod d}

Nilpotent shift: A = I + N

Binomial for a nilpotent shift

A = I + N,\ N^k = O \ \Rightarrow\ A^n = I + nN + \binom{n}{2}N^2 + \dots

$N$ nilpotent part A − I (strictly triangular)
$\binom{n}{2}$ n(n−1)/2, the coefficient of N²

Idempotent (A² = A) and involutory (A² = I) matrices

Idempotent binomial collapse

A^2 = A \ \Rightarrow\ (I+A)^n = I + (2^n - 1)A

The Cayley-Hamilton equation

Cayley-Hamilton (2×2)

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

$\operatorname{tr}A$ a + d, sum of diagonal entries
$\det A$ ad − bc

Reducing higher powers with a polynomial relation

Power reduction

A^2 = pA + qI \ \Rightarrow\ A^3 = (p^2+q)A + pq\,I

Sums of powers of a matrix

Geometric sum for a rank-1 matrix

A^2 = cA \ \Rightarrow\ \sum_{k=1}^{m}A^k = \frac{c^m - 1}{c - 1}\,A

Powers under conjugation: (P⁻¹BP)ⁿ

Conjugation commutes with powers

(P^{-1}BP)^n = P^{-1}B^{n}P \qquad (P^{T}P = I \Rightarrow P^{T}(PBP^{T})^{n}P = B^{n})

Counting n with Aⁿ = A or Aⁿ = I

Counting the exponents

A^n = I \iff d \mid n \ \Rightarrow\ \#\{n \le N\} = \left\lfloor \tfrac{N}{d} \right\rfloor

Eigenvalue and trace-determinant reasoning

Sum and product of eigenvalues

\lambda_1 + \lambda_2 = \operatorname{tr}A,\qquad \lambda_1\lambda_2 = \det A

Common traps

Reduce the exponent modulo the period FIRST

Once you find

A^d = I

, a power like

A^{2019}

is just

A^{\,2019 \bmod d}

. Students who start multiplying

A\cdot A\cdot A\dots

never finish. Find the period, take the remainder, done.

The binomial TRUNCATES — don't chase infinitely many terms

For

A = I + N

, the expansion of

A^n

stops at the last non-zero

N^k

. A

3\times3

unitriangular matrix has

N^3 = O

, so you need exactly three terms — no more. Writing an endless series is wasted effort.

This is not a normal binomial — $I$ and $N$ must commute

(I+N)^n = \sum \binom{n}{k}N^k

is valid ONLY because

I

commutes with

N

. You may never apply the binomial theorem to

(A+B)^n

for two general matrices; here it works purely because one summand is

I

$(I+A)^n = I + (2^n-1)A$ needs $A^2 = A$

This clean collapse holds ONLY for an idempotent

A

. If

A^2 \ne A

, the binomial doesn't fold up — you must use the actual structure (nilpotent, cyclic, or Cayley-Hamilton). Always verify

A^2 = A

before you use it.

The determinant term is $+\det A$ for a $2\times2$

Cayley-Hamilton is

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

: the trace term is negative, the determinant term is positive. Flipping the sign of

\det A

is the single most common slip — read the constant term off directly as

+\det A

Cayley-Hamilton is about the CHARACTERISTIC polynomial

The matrix satisfies

p(A) = O

where

p(\lambda) = \det(A - \lambda I)

— not some arbitrary polynomial you'd like. For

3\times3

the coefficients are

\operatorname{tr}A

, the sum of principal

2\times2

minors, and

\det A

Substitute the relation at every step — don't expand entrywise

When a problem hands you

A^2 = pA + qI

, you are meant to REDUCE powers algebraically, not compute the matrix. Re-apply the relation each time a fresh

A^2

appears; the answer always lands as

\alpha A + \beta I

Spot the rank-1 structure before you multiply

If a matrix is a column times a row (e.g.

a_{ij} = 2^{\,j-i} = 2^{-i}\cdot 2^{j}

), then

A^2 = (\operatorname{tr}A)A

and every power is a scalar times

A

. Recognising this turns a scary

\sum A^k

into a plain geometric series.

Strip the conjugation FIRST

(P^{-1}BP)^n

is never expanded directly. The middle

P P^{-1}

pairs telescope, leaving

P^{-1}B^{n}P

. For orthogonal

P

remember

P^{-1} = P^{T}

, so

P^{T}(PBP^{T})^{n}P

collapses all the way to

B^{n}

Find the PERIOD, then count residues — don't test each n

The whole question reduces to "how many

n \le N

hit the right residue mod

d

". Nail the period first (often the order of a complex eigenvalue like

i

\omega

), and the rest is arithmetic. Testing powers one by one is a trap for time.

$\operatorname{tr}(A^2) \ne (\operatorname{tr}A)^2$

The sum of diagonal entries of

A^2

\lambda_1^2 + \lambda_2^2 = (\operatorname{tr}A)^2 - 2\det A

, NOT

(\operatorname{tr}A)^2

. Forgetting the

-2\det A

correction is the standard error on these trace questions.

Symmetric, Skew-Symmetric and Orthogonal Matrices

Learn this subtopic in the notes

Definitions, entry patterns, and counting

Free-entry counts for order n

\#\{\text{symmetric}\} = k^{\,n(n+1)/2}, \qquad \#\{\text{skew}\} = \begin{cases} k^{\,n(n-1)/2} & 0 \in \text{value set}\\[2pt] 0 & 0 \notin \text{value set}\end{cases}

$n$ order of the square matrix
$k$ number of allowed values per entry

The symmetric-plus-skew decomposition

Unique symmetric + skew split

A = \underbrace{\tfrac12(A+A^T)}_{P\ \text{symmetric}} + \underbrace{\tfrac12(A-A^T)}_{Q\ \text{skew}}

Transposing products of symmetric and skew matrices

Transpose rules and skew-power parity

(XY)^T = Y^T X^T, \qquad (B^n)^T = (-1)^n B^n \ \ (B\ \text{skew})

The quadratic form test for skew-symmetry

Quadratic-form characterisation

X^TAX = 0 \ \ \forall X \;\Longrightarrow\; A^T = -A \ \ (A\ \text{skew-symmetric})

Orthogonal, rotation, and Cayley-transform matrices

Orthogonality and rotation composition

AA^T=I \;\Rightarrow\; A^{-1}=A^T,\ \det A=\pm1 \qquad f(x)f(y)=f(x+y)

Common traps

A symmetric matrix is NOT determined by $n^2$ free choices

For a symmetric

3\times3

matrix you choose only

\tfrac{3\cdot4}{2}=6

entries (3 diagonal + 3 upper), not

9

. Counting all

9

entries freely over-counts massively — the lower triangle is a mirror, not a free choice.

Skew-symmetric over a set missing $0$ gives ZERO matrices

The diagonal of a skew matrix is forced to

0

. If the allowed value set does not contain

0

(e.g.

\{-3,-2,-1,1,2\}

), the skew count collapses to

0

— a favourite JEE curveball hidden inside a union count.

$B^n$ flips type with the parity of $n$

For skew

B

B^{26}

is symmetric but

B^{5}

is skew. Students who assume every power of a skew matrix stays skew mis-answer questions like

B^{5}-A^{5}

: since

(B^5)^T=-B^5

and

(A^5)^T=A^5

, the difference is neither symmetric nor skew in general.

A product of symmetric matrices need not be symmetric

(AB)^T=B^TA^T=BA

for symmetric

A,B

— which equals

AB

only if they commute. Don't declare

AB

symmetric automatically; it is symmetric iff

AB=BA

$X^TAX=0$ does not mean $A=O$

The vanishing quadratic form only annihilates the symmetric part

A+A^T

. A non-zero skew matrix such as

\begin{pmatrix}0&2\\-2&0\end{pmatrix}

satisfies

X^TAX=0

identically while being far from the zero matrix. Read the condition as 'skew', not 'zero'.

$\det A=\pm1$ , not always $+1$

Orthogonality only pins

(\det A)^2=1

. Proper rotations have

\det=+1

, but reflections are orthogonal with

\det=-1

. Never assume the determinant is

1

unless the matrix is specifically a rotation.

Orthogonal is about $A^T=A^{-1}$ , not $A^T=A$

Don't confuse orthogonal (

AA^T=I

, so

A^T=A^{-1}

) with symmetric (

A^T=A

). A symmetric matrix with

A^2=I

happens to be orthogonal too, but symmetry alone never implies orthogonality.

Adjoint, Inverse & Determinant Identities

Learn this subtopic in the notes

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)

Common traps

$\det(kA) = k^{\,n}\det A$ , 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.

$|\operatorname{adj}A| = |A|^{\,n-1}$ — 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.

$\operatorname{adj}(\operatorname{adj}A) = |A|^{\,n-2}A$ : 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.

$PQ = kI_n \Rightarrow |Q| = k^{\,n}/|P|$ — 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.

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

$MN = O$ 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.

$A = A^{-1}$ 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.

Matrices formulas

Matrix Algebra, Types & Operations

Order, equality, linear combinations, and trace

Matrix multiplication: conformability and the row-by-column rule

Transpose and the reversal law

Non-commutativity, zero divisors, and commuting matrices

Elementary row operations

Counting matrices

Common traps

Powers of a Matrix & the Cayley-Hamilton Theorem

Cyclic and pattern powers

Nilpotent shift: A = I + N

Idempotent (A² = A) and involutory (A² = I) matrices

The Cayley-Hamilton equation

Reducing higher powers with a polynomial relation

Sums of powers of a matrix

Powers under conjugation: (P⁻¹BP)ⁿ

Counting n with Aⁿ = A or Aⁿ = I

Eigenvalue and trace-determinant reasoning

Common traps

Symmetric, Skew-Symmetric and Orthogonal Matrices

Definitions, entry patterns, and counting

The symmetric-plus-skew decomposition

Transposing products of symmetric and skew matrices

The quadratic form test for skew-symmetry

Orthogonal, rotation, and Cayley-transform matrices

Common traps

Adjoint, Inverse & Determinant Identities

Determinant of products, transposes and scalar multiples

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

Adjoint of the adjoint

Inverse from the adjoint and PQ = kI

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

When is a matrix invertible? (det ≠ 0)

Special inverses, counting, and the reversal law

Common traps

More JEE Mains Maths formula sheets