---
title: "Cayley-Hamilton theorem verification [4,3,1;2,1,−2;…]"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:cayley-hamilton-theorem-verification-4-3"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/cayley-hamilton-theorem-verification-4-3
site: BCT Engineering
---

# Cayley-Hamilton theorem verification [4,3,1;2,1,−2;…]

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2073 Chaitra, 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-4-3
- 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 [4 3 1; 2 1 −2; 1 2 1].

## Short answer

Cayley-Hamilton theorem: every square matrix satisfies its own characteristic equation. For the given matrix, |A|=11 and A³−6A²+6A−11I=0, so A⁻¹ = (1/11)(A²−6A+6I) = (1/11)[5 −1 −7; −4 3 10; 3 −5 −2].

## 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.

A = [4 3 1; 2 1 −2; 1 2 1]

#### Characteristic equation
- Trace = 4 + 1 + 1 = 6
- Sum of principal minors = (1·1 − (−2)·2) + (4·1 − 1·1) + (4·1 − 3·2) = 5 + 3 − 2 = 6
- |A| = 4(1 + 4) − 3(2 + 2) + 1(4 − 1) = 20 − 12 + 3 = 11

```text
λ³ − 6λ² + 6λ − 11 = 0
```

By the theorem: **A³ − 6A² + 6A − 11I = O**.

#### Inverse
|A| = 11 ≠ 0, so A⁻¹ exists. Multiply the relation by A⁻¹:

```text
A² − 6A + 6I − 11A⁻¹ = O   ⇒   A⁻¹ = (1/11)(A² − 6A + 6I)
```

Compute A²:

```text
A² = [4 3 1][4 3 1]   [23 17 −1]
     [2 1 −2][2 1 −2] = [ 8  3 −2]
     [1 2 1][1 2 1]   [ 9  7 −2]
```

Then

```text
A² − 6A + 6I = [23−24+6   17−18    −1−6 ]   [ 5 −1 −7]
               [ 8−12     3−6+6    −2+12]  = [−4  3 10]
               [ 9−6      7−12     −2−6+6]   [ 3 −5 −2]
```

**Answer: A⁻¹ = (1/11) [5 −1 −7; −4 3 10; 3 −5 −2].**

Check: row 1 of A times column 1 of the matrix: (4·5 + 3·(−4) + 1·3)/11 = (20 − 12 + 3)/11 = 1 ✓; row 1 times column 2: (−4 + 9 − 5)/11 = 0 ✓.

## Check yourself

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

1. Cayley–Hamilton says:
   A. Every matrix has an inverse
   B. Every square matrix satisfies its own characteristic equation
   C. Eigenvalues are real
   D. A² = A
   Answer: B. The matrix substituted into its characteristic polynomial gives the zero matrix. (Key point: Theorem)

2. Characteristic equation of the matrix:
   A. λ³ − 6λ² + 6λ − 11 = 0
   B. λ³ + 6λ² + 6λ + 11 = 0
   C. λ³ − 6λ² − 6λ − 11 = 0
   D. λ³ − 11λ² + 6λ − 6 = 0
   Answer: A. Trace 6, minors sum 6, |A| = 11. (Key point: Characteristic equation)

3. The inverse is obtained as:
   A. A⁻¹ = (1/11)(A³ − 6A²)
   B. A⁻¹ = 11(A² − 6A)
   C. A⁻¹ = (A² + 6A + 6I)/11
   D. A⁻¹ = (1/11)(A² − 6A + 6I)
   Answer: D. Multiply A³ − 6A² + 6A − 11I = O by A⁻¹ and solve. (Key point: Inverse formula)

4. Why can we compute A⁻¹ this way?
   A. A is symmetric
   B. A is diagonal
   C. |A| = 11 ≠ 0
   D. Trace is 6
   Answer: C. The inverse exists only because the determinant is non-zero. (Key point: Condition)

## Close variants set in other exams

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