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

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

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2070 Asar, 5 marks
- 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-2-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Verify Cayley-Hamilton theorem for the matrix A = [2 −1 1; −1 2 −1; 1 −1 2] and find the inverse of A.

## Short answer

Cayley-Hamilton theorem holds: A³ − 6A² + 9A − 4I = O. Thus A⁻¹ = (1/4)(A² − 6A + 9I) = (1/4)[3 1 −1; 1 3 1; −1 1 3].

## Answer

A = [2 −1 1; −1 2 −1; 1 −1 2]

#### Characteristic equation
- Trace = 2 + 2 + 2 = 6
- Sum of principal minors = (4 − 1) + (4 − 1) + (4 − 1) = 9
- |A| = 2(4 − 1) + 1(−2 + 1) + 1(1 − 2) = 6 − 1 − 1 = 4

```text
λ³ − 6λ² + 9λ − 4 = 0      [ = (λ − 1)²(λ − 4) ]
```

#### Verification: A³ − 6A² + 9A − 4I = O
```text
A² = [ 6 −5  5]         A³ = [ 22 −21  21]
     [−5  6 −5]              [−21  22 −21]
     [ 5 −5  6]              [ 21 −21  22]
```

```text
Row 1: 22 − 36 + 18 − 4 = 0 ;  −21 + 30 − 9 = 0 ;  21 − 30 + 9 = 0
Row 2: −21 + 30 − 9 = 0 ;  22 − 36 + 18 − 4 = 0 ;  −21 + 30 − 9 = 0
Row 3: 21 − 30 + 9 = 0 ;  −21 + 30 − 9 = 0 ;  22 − 36 + 18 − 4 = 0
```

All entries are zero, so Cayley–Hamilton is verified. ∎

#### Inverse
Since |A| = 4 ≠ 0, multiply A³ − 6A² + 9A − 4I = O by A⁻¹:

```text
A² − 6A + 9I − 4A⁻¹ = O   ⇒   A⁻¹ = (1/4)(A² − 6A + 9I)
```

```text
A² − 6A + 9I = [ 6−12+9   −5+6    5−6  ]   [ 3  1 −1]
               [ −5+6     6−12+9  −5+6 ] = [ 1  3  1]
               [ 5−6      −5+6    6−12+9]   [−1  1  3]
```

**Answer: A⁻¹ = (1/4) [3 1 −1; 1 3 1; −1 1 3].**

Check: row 1 of A times column 1: (2·3 − 1·1 + 1·(−1))/4 = 4/4 = 1 ✓.

## Check yourself

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

1. Equation for the matrix:
   A. λ³ − 6λ² + 4λ − 9 = 0
   B. λ³ − 6λ² + 9λ − 4 = 0
   C. λ³ + 6λ² + 9λ + 4 = 0
   D. λ³ − 9λ² + 6λ − 4 = 0
   Answer: B. Trace 6, minors sum 9, det 4. (Key point: Characteristic equation)

2. To verify the theorem we show:
   A. A³ − 6A² + 9A − 4I = O
   B. A³ = I
   C. A² = 6A
   D. |A| = 0
   Answer: A. Substituting A gives the zero matrix. (Key point: Verification)

3. Inverse formula:
   A. A⁻¹ = 4(A² − 6A + 9I)
   B. A⁻¹ = (1/9)(A² − 6A + 4I)
   C. A⁻¹ = (1/4)(A³ − 6A²)
   D. A⁻¹ = (1/4)(A² − 6A + 9I)
   Answer: D. Multiply by A⁻¹ and divide by 4. (Key point: Inverse)

4. A⁻¹ exists because:
   A. trace ≠ 0
   B. A is symmetric only
   C. |A| = 4 ≠ 0
   D. eigenvalues are all equal
   Answer: C. Non-zero determinant is the condition. (Key point: Why inverse exists)

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