JEE Mains Maths · Matrices

Powers of a Matrix & the Cayley-Hamilton Theorem

To find a high power of a matrix you almost never multiply it out — you spot a structure (a cycle, a nilpotent shift, an idempotent, or the matrix's own characteristic equation) that collapses every power into a simple pattern.

All Matrices notes JEE Mains Maths strategy

Why this matters

Fifty PYQs, one from almost every JEE Mains sitting 2021-2026 — the single most-tested idea in the Matrices chapter, and all MODERATE. Nobody computes A raised to 2007 by hand: the exam rewards recognising WHICH shortcut applies. The nine tools below — cyclic powers, the nilpotent A = I + N binomial, idempotent/involutory reductions, the Cayley-Hamilton equation, polynomial recurrences, sums of powers, conjugation P^{-1}BP, period-counting, and eigenvalue/trace-determinant reasoning — between them crack every one of these 50 questions in under two minutes.

Concept 1 of 9: Cyclic and pattern powers

Compute

A^2

and

A^3

and one of two things happens: the entries follow an obvious pattern (e.g.

A^n

has

n

n-1

in one slot), or the matrix cycles —

A^d = I

for some small period

d

, so

A^n

depends only on

n \bmod d

. Rotations and permutation matrices are the classic cyclers.

Definition

For a square matrix $A$ :

Pattern powers: find $A^2, A^3$ , guess the closed form of $A^n$ , and (optionally) confirm by induction.
Cyclic powers: if $A^d = I$ (smallest such $d$ is the period), then $A^n = A^{\,n \bmod d}$ . A rotation by $\theta$ satisfies $A^n =$ rotation by $n\theta$ ; a rotation by $2\pi/d$ has period $d$ .
Reduce the exponent FIRST, then read off the answer.

Cyclic reduction

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

Worked example

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

, find

A^{2019}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q158Moderate

Example 1 · Matrices · Powers of a Matrix and Cayley-Hamilton Theorem

A =\frac{1}{2}\begin{bmatrix} 1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{bmatrix}

, then:

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.

Drill 6 more on cyclic and pattern powers

Concept 2 of 9: Nilpotent shift: A = I + N

A triangular matrix with 1's on the diagonal is

I

plus a nilpotent part

N

(its powers eventually vanish:

N^k = O

). Because

I

commutes with everything, the binomial theorem applies to

(I+N)^n

— and it truncates, because every term with

N^k = O

drops out. So

A^n

is a short, exact polynomial in

N

Definition

Write $A = I + N$ where $N$ is nilpotent ( $N^k = O$ for some $k$ ). Then

A^n = (I+N)^n = I + nN + \binom{n}{2}N^2 + \dots + \binom{n}{k-1}N^{k-1},

the series stopping at the last non-zero power of

N

. For a

2\times2

unitriangular matrix

N^2 = O

, so

A^n = I + nN

; for a

3\times3

unitriangular matrix

N^3 = O

, so

A^n = I + nN + \binom{n}{2}N^2

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

Worked example

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

, find

A^4

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 25 July 2022 · Q173Moderate

Example 2 · Matrices · Powers of a Matrix and Cayley-Hamilton Theorem

Let

A =\begin{bmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{bmatrix},a,b \in R

. If for some

n \in N

A^{n}=\begin{bmatrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{bmatrix}

then

n + a + b

is equal to

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

Drill 6 more on nilpotent shift: a = i + n

Concept 3 of 9: Idempotent (A² = A) and involutory (A² = I) matrices

Two special self-relations make every power trivial. If

A^2 = A

(idempotent), then

A^3 = A^2\cdot A = A

, and so on:

A^n = A

for all

n \ge 1

. If

A^2 = I

(involutory), powers just alternate between

A

and

I

. Spot one of these and any expression in powers of

A

collapses instantly.

Definition

Idempotent: $A^2 = A \Rightarrow A^n = A$ for every $n \ge 1$ .
Involutory: $A^2 = I \Rightarrow A^n = I$ (n even), $A^n = A$ (n odd).
Binomial payoff (idempotent): $(I+A)^n = \sum_{k=0}^{n}\binom{n}{k}A^k = I + \Big(\sum_{k=1}^{n}\binom{n}{k}\Big)A = I + (2^n - 1)A$ , since every $A^k = A$ .

Idempotent binomial collapse

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

Worked example

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

, find

(I + A)^4

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 2 Apr 2025 · Q55Moderate

Example 3 · Matrices · Powers of a Matrix and Cayley-Hamilton Theorem

Let

A=\begin{bmatrix} \alpha & - 1 \\ 6 & \beta \end{bmatrix},\alpha> 0

, such that

det(A) = 0

and

\alpha+\beta= 1

. If I denotes

2 \times 2

identity matrix, then the matrix

(1 +A)^{8}

is:

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

Drill 4 more on idempotent (a² = a) and involutory (a² = i) matrices

Concept 4 of 9: The Cayley-Hamilton equation

Every square matrix satisfies its own characteristic equation. For a

2\times2

matrix that reads

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

— a single relation that instantly gives the trace, the determinant, or an unknown constant in an equation like

A^2 + \gamma A + cI = O

. Matching your given relation against Cayley-Hamilton is often a one-line solve.

Definition

2×2: $A^2 - (\operatorname{tr}A)A + (\det A)I = O$ , i.e. $A^2 - (a+d)A + (ad-bc)I = O$ . 3×3: $A^3 - (\operatorname{tr}A)A^2 + (M)A - (\det A)I = O$ , where $M$ is the sum of the three $2\times2$ principal minors. To solve "find $k$ in $A^2 - kA + cI = O$ ", just compare coefficients: $k = \operatorname{tr}A$ , $c = \det A$ . The sign on the determinant term is $+\det A$ for $2\times2$ .

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

Worked example

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

and

A^2 + pA + qI = O

, find

p

and

q

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 July 2022 · Q154Moderate

Example 4 · Matrices · Powers of a Matrix and Cayley-Hamilton Theorem

Let

A =\begin{pmatrix} 4 & - 2 \\ \alpha & \beta \end{pmatrix}

A^{2}+ \gamma A + 18I = O

, then

det(a)

is equal to

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

Drill 5 more on the cayley-hamilton equation

Concept 5 of 9: Reducing higher powers with a polynomial relation

Once you know a relation like

A^2 = pA + qI

(from Cayley-Hamilton or given outright), EVERY higher power is a linear combination of

A

and

I

: multiply the relation by

A

and substitute repeatedly.

A^3, A^4, A^5, \dots

all reduce to the form

\alpha A + \beta I

— you never need the actual entries.

Definition

Given $A^2 = pA + qI$ :

$A^3 = A\cdot A^2 = pA^2 + qA = (p^2+q)A + pq\,I$ , and so on — each step re-uses $A^2 = pA + qI$ .
The same works for a cubic relation $A^3 = pA^2 + qA + rI$ : express $A^4, A^5$ in terms of $A^2, A, I$ .
Trace shortcut: if $\lambda^2 = p\lambda + q$ is the characteristic equation, the power sums $s_k = \operatorname{tr}(A^k)$ obey the same recurrence $s_k = p\,s_{k-1} + q\,s_{k-2}$ , with $s_0 = 2,\ s_1 = \operatorname{tr}A$ .

Power reduction

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

Worked example

A matrix satisfies

A^2 = 3A + 2I

. Express

A^4

in the form

\alpha A + \beta I

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q61Moderate

Example 5 · Matrices · Powers of a Matrix and Cayley-Hamilton Theorem

Let

\alpha

and

\beta

be real numbers. Consider a

3 \times 3

matrix A such that

A^{2}= 3A + \alpha I

. If

A^{4}= 21A + \beta I

, then

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

Drill 7 more on reducing higher powers with a polynomial relation

Concept 6 of 9: Sums of powers of a matrix

A series

A + A^2 + \dots + A^m

is easy the moment each power is a scalar multiple of a fixed matrix. Two structures do that: a rank-1 matrix (a column times a row) satisfies

A^2 = cA

, so

A^k = c^{k-1}A

— a geometric series in

A

; and a matrix with

M^2 = cI

makes every even power a scalar multiple of

I

. Unitriangular sums add up term-by-term via the

I+N

pattern.

Definition

Rank-1 $A = uv^{T}$ : $A^2 = (v^{T}u)A = cA$ , hence $A^k = c^{k-1}A$ and $\sum_{k=1}^{m}A^k = \dfrac{c^m - 1}{c - 1}\,A$ (for $c \ne 1$ ). Here $c = \operatorname{tr}A$ .
$M^2 = cI$ : even powers $M^{2k} = c^{k}I$ , odd powers $M^{2k-1} = c^{k-1}M$ — split the sum into even and odd parts.
Unitriangular $A = I + N$ : $\sum A^k = (\sum 1)I + (\sum k)N + (\sum \binom{k}{2})N^2$ , summing each coefficient separately.

Geometric sum for a rank-1 matrix

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

Worked example

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

, find

A + A^2 + A^3 + A^4 + A^5

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 June 2022 · Q70Moderate

Example 6 · Matrices · Powers of a Matrix and Cayley-Hamilton Theorem

Let

A =\left\lbrack a_{ij} \right\rbrack

be a square matrix of order 3 such that

a_{ij}=2^{j - i}

, for all

i,j = 1,2,3

. Then, the matrix

A^{2}+A^{3}+ \ldots +A^{10}

is equal to:

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.

Drill 3 more on sums of powers of a matrix

Concept 7 of 9: Powers under conjugation: (P⁻¹BP)ⁿ

When a matrix is dressed up as

P^{-1}BP

(a similarity/conjugation), its powers un-dress cleanly: the inner

P

's cancel, so

(P^{-1}BP)^n = P^{-1}B^{n}P

. For an orthogonal

P

(

P^{T}P = I

) the inverse is just

P^{T}

. This is the whole trick behind the "

P^{T}Q^{2007}P

" monsters — strip the conjugation, power the simple middle matrix, done.

Definition

For invertible $P$ : $(P^{-1}BP)^n = P^{-1}B^{n}P$ , because $(P^{-1}BP)(P^{-1}BP) = P^{-1}B(PP^{-1})BP = P^{-1}B^{2}P$ , and so on by induction.

If $P$ is orthogonal ( $P^{T}P = I$ ), then $P^{-1} = P^{T}$ , so $Q = PBP^{T} \Rightarrow P^{T}Q^{n}P = B^{n}$ .
Similar matrices share trace, determinant, and eigenvalues, so a diagonal-element sum of $P^{T}Q^{n}P$ is just that of $B^{n}$ .

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})

Worked example

Let

P

be orthogonal (

P^{T}P = I

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

, and

Q = PBP^{T}

. Find

P^{T}Q^{5}P

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 8 Apr 2023 · Q65Moderate

Example 7 · Matrices · Powers of a Matrix and Cayley-Hamilton Theorem

Let

P =\begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix},A =\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}

And

Q =PQP^{T}

. If

\ P^{T}Q^{2007}P =\begin{bmatrix} a & b \\ c & d \end{bmatrix}

, then

2a + b - 3c - 4d

equal to

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}

Drill 4 more on powers under conjugation: (p⁻¹bp)ⁿ

Concept 8 of 9: Counting n with Aⁿ = A or Aⁿ = I

"For how many

n \in \{1, \dots, N\}

A^n = I

?" is really a modular-counting question. Find the period

d

(smallest

d

with

A^d = I

); then

A^n = I \iff d \mid n

, and you just count the multiples of

d

up to

N

. For

A^n = A

, the condition is

n \equiv 1 \pmod{d}

Definition

Determine the period $d$ (via cyclicity, eigenvalues like $i$ or $\omega$ , or a relation such as $A^2 = A$ ).
$A^n = I \iff n \equiv 0 \pmod d$ ; count $= \lfloor N/d \rfloor$ .
$A^n = A \iff n \equiv 1 \pmod d$ .
When eigenvalues are complex roots of unity ( $i$ has order 4, $\omega$ order 3), the order is the period. If $A^2 = A$ (idempotent), $A^n = A$ for ALL $n \ge 1$ .

Counting the exponents

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

Worked example

A matrix has period 4 (

A^4 = I

, no smaller). For how many

n \in \{1, 2, \dots, 50\}

A^n = I

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 28 June 2022 · Q177Moderate

Example 8 · Matrices · Powers of a Matrix and Cayley-Hamilton Theorem

Let

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

where

i =\sqrt{- 1}

. Then, the number of elements in the set

\left\{ \mathbf{n \in \{ 1,2,\ldots,100\}:}A^{n}= A \right\}

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.

Drill 3 more on counting n with aⁿ = a or aⁿ = i

Concept 9 of 9: Eigenvalue and trace-determinant reasoning

The roots of

\det(A - \lambda I) = 0

(the eigenvalues) carry the matrix's DNA: their sum is the trace, their product the determinant. That lets you answer questions about

\operatorname{tr}(A^2)

\det A

, or hidden eigenvalues without ever finding the entries — and

AB = B

with

B \ne O

instantly tells you

1

is an eigenvalue.

Definition

For a $2\times2$ matrix $A$ with eigenvalues $\lambda_1, \lambda_2$ (the roots of $|A - \lambda I| = 0$ ):

$\lambda_1 + \lambda_2 = \operatorname{tr}A$ and $\lambda_1 \lambda_2 = \det A$ .
$\operatorname{tr}(A^2) = \lambda_1^2 + \lambda_2^2 = (\operatorname{tr}A)^2 - 2\det A$ .
If $AB = \lambda B$ for a non-zero column $B$ , then $\lambda$ is an eigenvalue of $A$ ; in particular $AB = B$ means $1$ is an eigenvalue.

Sum and product of eigenvalues

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

Worked example

For a

2\times2

matrix

A

, the roots of

|A - \lambda I| = 0

are

2

and

5

. Find

\operatorname{tr}(A^2)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q173Moderate

Example 9 · Matrices · Powers of a Matrix and Cayley-Hamilton Theorem

Let A be a

2 \times 2

real matrix and I be the identity matrix of order 2 . If the roots of the equation

|A-xI| = 0

be -1 and 3 , then the sum of the diagonal elements of the matrix

A^{2}

is.........

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

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.

Drill 3 more on eigenvalue and trace-determinant reasoning

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 (9)

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
$A = I + N,\ N^k = O \ \Rightarrow\ A^n = I + nN + \binom{n}{2}N^2 + \dots$
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)
$A^2 - (\operatorname{tr}A)\,A + (\det A)\,I = O$
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
$\lambda_1 + \lambda_2 = \operatorname{tr}A,\qquad \lambda_1\lambda_2 = \det A$

Watch out for (11)

Reduce the exponent modulo the period FIRST
→ Cyclic and pattern powers
The binomial TRUNCATES — don't chase infinitely many terms
→ Nilpotent shift: A = I + N
This is not a normal binomial — $I$ and $N$ must commute
→ Nilpotent shift: A = I + N
$(I+A)^n = I + (2^n-1)A$ needs $A^2 = A$
→ Idempotent (A² = A) and involutory (A² = I) matrices
The determinant term is $+\det A$ for a $2\times2$
→ The Cayley-Hamilton equation
Cayley-Hamilton is about the CHARACTERISTIC polynomial
→ The Cayley-Hamilton equation
Substitute the relation at every step — don't expand entrywise
→ Reducing higher powers with a polynomial relation
Spot the rank-1 structure before you multiply
→ Sums of powers of a matrix
Strip the conjugation FIRST
→ Powers under conjugation: (P⁻¹BP)ⁿ
Find the PERIOD, then count residues — don't test each n
→ Counting n with Aⁿ = A or Aⁿ = I
$\operatorname{tr}(A^2) \ne (\operatorname{tr}A)^2$
→ Eigenvalue and trace-determinant reasoning

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