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

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

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2079 Bhadra, 1+4 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-1-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

State Cayley-Hamilton theorem. Use it to find the inverse of the matrix [1 1 3; 1 3 −3; 2 −4 −4].

## Short answer

Cayley–Hamilton: every square matrix satisfies its characteristic equation. Here λ³ − 32λ + 56 = 0, so A³ − 32A + 56I = O; multiplying by A⁻¹ gives A⁻¹ = (32I − A²)/56. The inverse is (1/56)[24 8 12; 2 10 −6; 10 −6 −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. It can be used to find A⁻¹ by multiplying the relation by A⁻¹.

A = [1 1 3; 1 3 −3; 2 −4 −4]

#### Characteristic equation
- Trace = 1 + 3 − 4 = 0
- Sum of principal minors = (3·(−4) − (−3)(−4)) + (1·(−4) − 3·2) + (1·3 − 1·1) = −24 − 10 + 2 = −32
- |A| = 1(−12 − 12) − 1(−4 + 6) + 3(−4 − 6) = −24 − 2 − 30 = −56

```text
λ³ − 0·λ² − 32λ + 56 = 0   i.e.   λ³ − 32λ + 56 = 0
```

#### Apply the theorem
```text
A³ − 32A + 56I = O
```
Since |A| = −56 ≠ 0, A⁻¹ exists. Multiply by A⁻¹:

```text
A² − 32I + 56A⁻¹ = O   ⇒   A⁻¹ = (1/56)(32I − A²)
```

#### Compute A²
```text
A² = [ 8  −8 −12]
     [−2  22   6]
     [−10  6  34]
```

#### Inverse
```text
32I − A² = [ 24   8  12]
           [  2  10  −6]
           [ 10  −6  −2]

A⁻¹ = (1/56) [24   8  12]   [3/7   1/7   3/14 ]
             [ 2  10  −6] = [1/28  5/28  −3/28]
             [10  −6  −2]   [5/28 −3/28  −1/28]
```

**Check:** first row of A times first column of A⁻¹: (24 + 2 + 30)/56 = 1 ✓; first row times second column: (8 + 10 − 18)/56 = 0 ✓.

**Answer: A⁻¹ = (1/56)[24 8 12; 2 10 −6; 10 −6 −2].**

## 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 square matrix has real eigenvalues
   C. Every square matrix is invertible
   D. The determinant of a matrix equals the sum of its eigenvalues
   Answer: A. Substituting A for λ in the characteristic polynomial gives the zero matrix. (Key point: Cayley-Hamilton statement)

2. For this A, the trace is 0, the sum of principal minors is -32 and |A| = -56. Which characteristic equation results?
   A. λ^3 - 32λ - 56 = 0
   B. λ^3 + 32λ + 56 = 0
   C. λ^3 - 32λ + 56 = 0
   D. λ^3 - 56λ^2 - 32λ = 0
   Answer: C. The equation is λ^3 - (trace)λ^2 + (sum of minors)λ - |A| = λ^3 - 32λ + 56 = 0. (Key point: Characteristic equation coefficients)

3. From A^3 - 32A + 56I = O, how is A^-1 obtained?
   A. Multiply by A^-1 and get A^-1 = (1/56)(32I - A^2)
   B. Multiply by A^-1 and get A^-1 = (1/56)(A^2 - 32I)
   C. Divide by A and get A^-1 = 32A - 56I
   D. Take the transpose of A^2 and divide by 56
   Answer: A. A^2 - 32I + 56A^-1 = O, so 56A^-1 = 32I - A^2. (Key point: Finding A^-1)

4. Why can the inverse be found at all for this matrix?
   A. Because its trace is 0
   B. Because |A| = -56 is not zero
   C. Because A is symmetric
   D. Because A^2 is computed
   Answer: B. A matrix is invertible only if its determinant is non-zero. (Key point: Existence of inverse)

## Close variants set in other exams

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