---
title: "Cayley-Hamilton theorem verification [−2,2,−3;2,1,−6;…]"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:cayley-hamilton-theorem-verification-2-2-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/cayley-hamilton-theorem-verification-2-2-two
site: BCT Engineering
---

# Cayley-Hamilton theorem verification [−2,2,−3;2,1,−6;…]

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2079 Baishakh, 5 marks
- Video: [Cayley Hamilton : Inverse of a matrix | Step wise explanation | Example solved-1| Mathspedia |](https://www.youtube.com/watch?v=qJ-zw_meMQ0) — Mathspedia
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/cayley-hamilton-theorem-verification-2-2-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

State Cayley-Hamilton theorem and use it to find the inverse of the matrix [−2 2 −3; 2 1 −6; −1 −2 0].

## Short answer

The theorem states every square matrix satisfies its own characteristic equation. Here A³ + A² − 21A − 45I = O, giving A⁻¹ = (1/45)[−12 6 −9; 6 −3 −18; −3 −6 −6].

## Answer

**Cayley–Hamilton theorem:** every square matrix satisfies its own characteristic equation. If |A − λI| = λ³ + c₁λ² + c₂λ + c₃ = 0, then A³ + c₁A² + c₂A + c₃I = O. It lets us write A⁻¹ as a polynomial in A.

A = [−2 2 −3; 2 1 −6; −1 −2 0]

#### Characteristic equation
- Trace = −1
- Sum of principal minors = −12 − 3 − 6 = −21
- |A| = 45

```text
λ³ + λ² − 21λ − 45 = 0
```

So **A³ + A² − 21A − 45I = O**.

#### Inverse
|A| = 45 ≠ 0, so A⁻¹ exists. Multiply by A⁻¹:

```text
A² + A − 21I − 45A⁻¹ = O   ⇒   A⁻¹ = (1/45)(A² + A − 21I)
```

Compute A²:

```text
A² = [−2 2 −3][−2 2 −3]   [11   4  −6]
     [ 2 1 −6][ 2 1 −6] = [ 4  17 −12]
     [−1 −2 0][−1 −2 0]   [−2  −4  15]
```

Then A² + A − 21I:

```text
[11−2−21    4+2     −6−3 ]   [−12   6  −9]
[ 4+2     17+1−21  −12−6 ] = [  6  −3 −18]
[−2−1      −4−2    15+0−21]   [ −3  −6  −6]
```

**Answer: A⁻¹ = (1/45) [−12 6 −9; 6 −3 −18; −3 −6 −6].**

Check: row 1 of A times column 1 of the matrix: (−2·(−12) + 2·6 + (−3)(−3))/45 = (24 + 12 + 9)/45 = 1 ✓; row 1 times column 2: (−12 − 6 + 18)/45 = 0 ✓.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Cayley–Hamilton states:
   A. A matrix equals its transpose
   B. A matrix satisfies its own characteristic equation
   C. Its determinant is 1
   D. Its eigenvalues are 0
   Answer: B. p(A) = O. (Key point: Theorem)

2. For this matrix the relation is:
   A. A³ + A² − 21A − 45I = O
   B. A³ − A² − 21A + 45I = O
   C. A³ + A² + 21A + 45I = O
   D. A³ − 21A − 45I = O
   Answer: A. Characteristic equation λ³+λ²−21λ−45 = 0. (Key point: Relation)

3. Inverse formula:
   A. A⁻¹ = 45(A² + A − 21I)
   B. A⁻¹ = (1/21)(A² + A − 45I)
   C. A⁻¹ = (1/45)(A³ + A²)
   D. A⁻¹ = (1/45)(A² + A − 21I)
   Answer: D. Multiply by A⁻¹ and divide by 45. (Key point: Inverse)

4. Cayley–Hamilton gives A⁻¹ as:
   A. A product of eigenvalues
   B. A transpose
   C. A polynomial in A of lower degree
   D. Identity plus A
   Answer: C. It expresses A⁻¹ using A², A, I. (Key point: Use)

## Close variants set in other exams

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