---
title: "Cayley-Hamilton theorem verification [1,0,−1;1,2,1;…]"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:cayley-hamilton-theorem-verification-1-0"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/cayley-hamilton-theorem-verification-1-0
site: BCT Engineering
---

# Cayley-Hamilton theorem verification [1,0,−1;1,2,1;…]

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2083 Baishakh, 1+4 marks; 2075 Chaitra, 1+4 marks (asked 2 times)
- Video: [Verify Cayley Hamilton for the matrix & find its inverse best example(PART-2)](https://www.youtube.com/watch?v=qLOaCyqSuVA) — EASY MATHS EASY TRICKS
- 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-1-0
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

State Cayley-Hamilton theorem and verify it for the matrix A = [1 0 −1; 1 2 1; 2 2 3].

## Short answer

Cayley–Hamilton theorem: every square matrix satisfies its own characteristic equation. For A, the characteristic equation is λ³−6λ²+11λ−6=0, and A³−6A²+11A−6I = O, verifying the theorem.

## Answer

**Cayley–Hamilton theorem:** every square matrix satisfies its own characteristic equation. If |A − λI| = λⁿ + c₁λⁿ⁻¹ + … + cₙ = 0, then Aⁿ + c₁Aⁿ⁻¹ + … + cₙI = O.

#### Characteristic equation
A = [1 0 −1; 1 2 1; 2 2 3]

- Sum of diagonal elements (trace) = 1 + 2 + 3 = 6
- Sum of principal 2 × 2 minors = (2·3 − 1·2) + (1·3 − (−1)·2) + (1·2 − 0·1) = 4 + 5 + 2 = 11
- |A| = 1(6 − 2) − 0 + (−1)(2 − 4) = 4 + 2 = 6

```text
λ³ − 6λ² + 11λ − 6 = 0
```

#### Verification: show A³ − 6A² + 11A − 6I = O
```text
A² = [−1 −2 −4]       A³ = A²·A = [−11 −12 −13]
     [ 5  6  4]                    [ 19  20  13]
     [10 10  9]                    [ 38  38  27]
```

Now compute:

```text
A³ − 6A² + 11A − 6I
 Row 1: (−11+6+11−6, −12+12+0−0, −13+24−11−0) = (0, 0, 0)
 Row 2: (19−30+11−0, 20−36+22−6, 13−24+11−0)   = (0, 0, 0)
 Row 3: (38−60+22−0, 38−60+22−0, 27−54+33−6)   = (0, 0, 0)
```

Every entry is zero, so A³ − 6A² + 11A − 6I = O.

Hence the Cayley–Hamilton theorem is verified for A. (The eigenvalues are 1, 2, 3.) ∎

## Check yourself

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

1. What does the Cayley-Hamilton theorem say?
   A. Every square matrix satisfies its own characteristic equation
   B. Every matrix is diagonalizable
   C. The trace equals the determinant
   D. Every matrix has real eigenvalues
   Answer: A. If p(lambda) = 0 is the characteristic equation then p(A) = O. (Key point: Statement)

2. For a 3x3 matrix, the characteristic equation is
   A. lambda^3 + (trace) lambda^2 + |A| = 0
   B. lambda^3 - (trace) lambda^2 + (sum of principal minors) lambda - |A| = 0
   C. lambda^3 - |A| lambda^2 + trace = 0
   D. lambda^2 - trace lambda + |A| = 0
   Answer: B. Coefficients are trace, sum of principal 2x2 minors and the determinant, with alternating signs. (Key point: Building the characteristic equation)

3. Which is the characteristic equation here (trace 6, minors 11, |A| 6)?
   A. lambda^3 + 6 lambda^2 + 11 lambda + 6 = 0
   B. lambda^3 - 11 lambda^2 + 6 lambda - 6 = 0
   C. lambda^3 - 6 lambda^2 + 11 lambda - 6 = 0
   D. lambda^3 - 6 lambda^2 - 11 lambda + 6 = 0
   Answer: C. Signs alternate: -trace, +minors, -determinant. (Key point: Characteristic equation)

4. What must be shown to verify the theorem?
   A. A^3 - 6A^2 + 11A - 6 = 0 as scalars
   B. A^2 = 6A
   C. |A| = 0
   D. A^3 - 6A^2 + 11A - 6I = O
   Answer: D. Replace lambda by A and the constant by a multiple of I. (Key point: Verification)

## Close variants set in other exams

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