MHT-CET Maths · Determinants and Matrices

Cayley–Hamilton, Matrix Polynomials and Powers

Every 2 × 2 matrix satisfies A² − (trace)A + |A|I = 0 — so A⁻¹ is a combination αI + βA, a factored polynomial in A gives A⁻¹ in one line, and powers of A cycle.

All Determinants and Matrices notes MHT-CET Maths strategy

Why this matters

17 PYQs at 35% HARD, and one stem — A⁻¹ = αI + βA for a given 2 × 2 matrix — has been set four times in three years, asked for α − β, α + β, x and y, or 2x + 3y. The rest of the page is the same theorem read differently: a matrix given by (A − 3I)(A − 5I) = 0, an inverse of A + 3I from A² − 4A + 3I = 0, and a power A²⁰²⁹ that collapses because A³ is a scalar. Nothing here needs the adjoint; that is the point of the page.

Concept 1 of 4: Cayley–Hamilton for 2 × 2: A² − (tr A)A + |A|·I = 0

A matrix satisfies its own characteristic equation. For

2 \times 2

that equation is

\lambda^2 - (\text{trace})\lambda + \det = 0

, so

A^2

is always a combination of

A

and

I

— which is why every higher power and the inverse reduce to

\alpha I + \beta A

Definition

Trace $\operatorname{tr}A = a + d$ ; determinant $|A| = ad - bc$ . Then $A^2 - (a + d)A + (ad - bc)I = O$ .
For $A = \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix}$ : trace $4$ , determinant $5$ , so $A^2 - 4A + 5I$ is the null matrix — no multiplication needed.
Consequences: $A^2 = (\operatorname{tr}A)A - |A|I$ ; multiply by $A^{-1}$ : $A = (\operatorname{tr}A)I - |A|A^{-1}$ , so $A^{-1} = \dfrac{(\operatorname{tr}A)I - A}{|A|}$ .
Verify the theorem on any $2 \times 2$ once by direct multiplication; after that, quote it.

Cayley–Hamilton (2 × 2)

A^2 - (\operatorname{tr}A)\,A + |A|\,I = O \qquad A^{-1} = \frac{(\operatorname{tr}A)\,I - A}{|A|}

$\operatorname{tr}A$ sum of the diagonal entries
$O$ the zero matrix

Worked example

For

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

, find

A^2 - 5A + 5I

without computing

A^2

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q117Moderate

Example 1 · Determinants and Matrices · Cayley–Hamilton, Matrix Polynomials and Powers

If matrix

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

, then

A^2 - 4A + 5I

is where

I

is a unit matrix.

Sign of the determinant term

It is

A^2 - (\operatorname{tr}A)A + |A|I

: minus the trace, PLUS the determinant. Writing

-|A|I

turns a null-matrix answer into a wrong 'symmetric matrix' option.

Concept 2 of 4: A⁻¹ = αI + βA: Read α and β From the Theorem

Rearranging Cayley–Hamilton gives

A^{-1} = \dfrac{\operatorname{tr}A}{|A|}I - \dfrac{1}{|A|}A

. So

\alpha = \dfrac{\operatorname{tr}A}{|A|}

and

\beta = -\dfrac{1}{|A|}

, with no adjoint and no entry-matching.

Definition

From $A^2 - tA + dI = O$ ( $t = \operatorname{tr}A$ , $d = |A|$ ): $A^{-1} = \dfrac{t}{d}I - \dfrac1d A$ . Hence $\alpha = \dfrac{t}{d}$ , $\beta = -\dfrac1d$ .
$A = \begin{pmatrix} 1 & 2 \\ -1 & 4 \end{pmatrix}$ : $t = 5$ , $d = 6$ , so $\alpha = \dfrac56$ , $\beta = -\dfrac16$ ; then $4(\alpha - \beta) = 4$ , $4(\alpha + \beta) = \dfrac83$ .
$A = \begin{pmatrix} 1 & 2 \\ -5 & 1 \end{pmatrix}$ with $A^{-1} = xA + yI$ : $t = 2$ , $d = 11$ , so $x = -\dfrac{1}{11}$ , $y = \dfrac{2}{11}$ , and $2x + 3y = \dfrac{4}{11}$ .
Cross-check by entry-matching if time allows: the $(1,2)$ entry of $\alpha I + \beta A$ is $2\beta$ , which must equal the $(1,2)$ entry of $\dfrac{\operatorname{adj}A}{|A|}$ .
Read what is asked: $\alpha - \beta$ , $\alpha + \beta$ , $(x, y)$ or a weighted sum — the same matrix has been asked all four ways.

Inverse as a combination

A^{-1} = \alpha I + \beta A \ \text{ with } \ \alpha = \frac{\operatorname{tr}A}{|A|},\ \beta = -\frac{1}{|A|}

Worked example

For

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

, write

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

and find

\alpha + \beta

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q127Moderate

Example 2 · Determinants and Matrices · Cayley–Hamilton, Matrix Polynomials and Powers

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

and

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

where

I

is the identity matrix of order 2 , hen

4(\alpha+\beta) =

α from the wrong entry

Matching the

(1,1)

entry gives

\alpha + \beta

, not

\alpha

— for

\begin{pmatrix} 1 & 2 \\ -1 & 4 \end{pmatrix}

that is

\frac23

, and reading it as

\alpha

makes

4(\alpha - \beta) = \frac{10}{3}

, which is offered.

\alpha = \operatorname{tr}A/|A| = \frac56

Drill 3 more on a⁻¹ = αi + βa: read α and β from the theorem

Concept 3 of 4: A Factored Polynomial in A Gives A⁻¹ in One Line

(A - 3I)(A - 5I) = O

expands to

A^2 - 8A + 15I = O

. Multiply through by

A^{-1}

and the inverse appears:

15A^{-1} = 8I - A

. No entries, no adjoint — the relation IS the inverse.

Definition

Expand the given factorisation (matrices with $I$ commute, so ordinary algebra applies): $(A - 3I)(A - 5I) = A^2 - 8A + 15I$ .
Multiply by $A^{-1}$ ( $A$ non-singular): $A - 8I + 15A^{-1} = O$ , so $A^{-1} = \dfrac{8I - A}{15}$ , i.e. $\dfrac{15}{8}A^{-1} = I - \dfrac18A$ ; and $\alpha A + \beta A^{-1} = 4I$ is matched by $\alpha = \dfrac12$ , $\beta = \dfrac{15}{2}$ , sum $8$ .
Inverse of a shifted matrix: from $A^2 - 4A + 3I = O$ , write $A^2 - 4A + 3I = (A + 3I)(A - 7I) + 24I$ , so $(A + 3I)(A - 7I) = -24I$ and $(A + 3I)^{-1} = \dfrac{7I - A}{24}$ . Choose the second factor so that the product's constant term matches.
The technique: treat $A$ like a number in polynomial identities, but never divide by a matrix — multiply by the inverse instead.

From a polynomial relation to the inverse

A^2 - pA + qI = O \ (q \ne 0) \ \Rightarrow\ A^{-1} = \frac{pI - A}{q} \qquad (A + kI)^{-1}:\ \text{write } A^2 - pA + qI = (A + kI)(A - mI) + cI

Worked example

A

is non-singular and

(A - 2I)(A - 4I) = O

, express

A^{-1}

in terms of

A

and

I

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q101Hard

Example 3 · Determinants and Matrices · Cayley–Hamilton, Matrix Polynomials and Powers

Suppose A is any

3\times3

non-singular matrix and

(A-3I)(A-5I)=O

where

I=I_3

and

O=O_3

. Here

O_3

represent zero matrix of order 3 and

I_3

is an identity matrix of order 3. If

\alpha A + \beta A^{-1} = 4I

, then

\alpha + \beta

is equal to

Concluding A = 3I or A = 5I

(A - 3I)(A - 5I) = O

does NOT force either factor to be zero — matrices have zero divisors. The information is the polynomial relation, and only that.

Drill 2 more on a factored polynomial in a gives a⁻¹ in one line

Concept 4 of 4: Powers of a Matrix: Find the Cycle

For

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

A^3

turns out to be

iI

, a scalar. Then

A^{12} = (iI)^4 = I

, and

A^{2029}

is just

A

— the exponent only matters modulo the cycle length.

Definition

Compute $A^2$ , $A^3$ , … until a scalar multiple of $I$ (or $I$ itself) appears. If $A^m = cI$ , then $A^{mk} = c^kI$ .
$A = \begin{pmatrix} i & 1 \\ 1 & 0 \end{pmatrix}$ : $A^2 = \begin{pmatrix} 0 & i \\ i & 1 \end{pmatrix}$ , $A^3 = iI$ , so $A^{12} = i^4I = I$ . $2029 = 12\cdot169 + 1$ , hence $A^{2029} = A$ .
The inverse of $A^{2029}$ is then $A^{-1} = \dfrac{\operatorname{adj}A}{|A|}$ with $|A| = -1$ : $-\operatorname{adj}A$ .
Cayley–Hamilton is the general engine: $A^2 = tA - dI$ lets any power be reduced to $pA + qI$ step by step; a cycle appears when $A^m$ lands on $cI$ .

Powers modulo a cycle

A^m = cI \ \Rightarrow\ A^{mk + r} = c^k A^r \qquad \begin{pmatrix} i & 1 \\ 1 & 0 \end{pmatrix}^3 = iI,\quad i^4 = 1

Worked example

For

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

, find

A^{101}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q144Hard

Example 4 · Determinants and Matrices · Cayley–Hamilton, Matrix Polynomials and Powers

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

where

i=\sqrt{-1}

and

B=A^{2029}

, then

B^{-1}=

Reducing 2029 modulo 3 instead of 12

A^3 = iI

is a scalar but NOT the identity; the cycle closes at

A^{12} = I

. Using

3

gives

A^{2029} = A^{1}\cdot i^{676} = A

only by luck of the arithmetic — reduce modulo the exponent at which the power returns to

I

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 (4)

Cayley–Hamilton for 2 × 2: A² − (tr A)A + |A|·I = 0
Cayley–Hamilton (2 × 2)
$A^2 - (\operatorname{tr}A)\,A + |A|\,I = O \qquad A^{-1} = \frac{(\operatorname{tr}A)\,I - A}{|A|}$
A⁻¹ = αI + βA: Read α and β From the Theorem
Inverse as a combination
$A^{-1} = \alpha I + \beta A \ \text{ with } \ \alpha = \frac{\operatorname{tr}A}{|A|},\ \beta = -\frac{1}{|A|}$
A Factored Polynomial in A Gives A⁻¹ in One Line
From a polynomial relation to the inverse
$A^2 - pA + qI = O \ (q \ne 0) \ \Rightarrow\ A^{-1} = \frac{pI - A}{q} \qquad (A + kI)^{-1}:\ \text{write } A^2 - pA + qI = (A + kI)(A - mI) + cI$
Powers of a Matrix: Find the Cycle
Powers modulo a cycle
$A^m = cI \ \Rightarrow\ A^{mk + r} = c^k A^r \qquad \begin{pmatrix} i & 1 \\ 1 & 0 \end{pmatrix}^3 = iI,\quad i^4 = 1$

Watch out for (4)

Sign of the determinant term
→ Cayley–Hamilton for 2 × 2: A² − (tr A)A + |A|·I = 0
α from the wrong entry
→ A⁻¹ = αI + βA: Read α and β From the Theorem
Concluding A = 3I or A = 5I
→ A Factored Polynomial in A Gives A⁻¹ in One Line
Reducing 2029 modulo 3 instead of 12
→ Powers of a Matrix: Find the Cycle

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