---
title: "Eigenvalues and eigenvectors [2,2,1;1,3,1;…]"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:eigenvalues-and-eigenvectors-2-2-1"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/eigenvalues-and-eigenvectors-2-2-1
site: BCT Engineering
---

# Eigenvalues and eigenvectors [2,2,1;1,3,1;…]

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2071 Chaitra, 5 marks; 2067 Asar (asked 2 times)
- Video: [3 x  3 eigenvalues and eigenvectors](https://www.youtube.com/watch?v=qa9fI6qvUQY) — Prime Newtons
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/eigenvalues-and-eigenvectors-2-2-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Find the eigenvalues and eigenvectors of the matrix [2 2 1; 1 3 1; 1 2 2].

## Short answer

Characteristic equation: λ³ − 7λ² + 11λ − 5 = 0 gives eigenvalues λ = 1 (twice) and λ = 5. For λ=5, eigenvector is (1,1,1)ᵀ. For λ=1, eigenvectors are (−2,1,0)ᵀ and (−1,0,1)ᵀ.

## Answer

A = [2 2 1; 1 3 1; 1 2 2]

#### Characteristic equation
- Trace = 2 + 3 + 2 = 7
- Sum of principal minors = (3·2 − 1·2) + (2·2 − 1·1) + (2·3 − 2·1) = 4 + 3 + 4 = 11
- |A| = 2(6 − 2) − 2(2 − 1) + 1(2 − 3) = 8 − 2 − 1 = 5

```text
λ³ − 7λ² + 11λ − 5 = 0   ⇒   (λ − 1)²(λ − 5) = 0
```

**Eigenvalues: λ = 1, 1, 5.**

#### Eigenvectors
**λ = 5:** A − 5I = [−3 2 1; 1 −2 1; 1 2 −3]. Adding R₁ and R₂ gives −2x + 2z = 0, so x = z. Putting x = z = 1 in R₂ gives 1 − 2y + 1 = 0, so y = 1.
X₃ = (1, 1, 1)ᵀ.

**λ = 1 (repeated):** A − I = [1 2 1; 1 2 1; 1 2 1]. All rows give the single equation x + 2y + z = 0, so rank = 1 and there are 3 − 1 = 2 independent eigenvectors.
- y = 1, z = 0 ⇒ x = −2: X₁ = (−2, 1, 0)ᵀ
- y = 0, z = 1 ⇒ x = −1: X₂ = (−1, 0, 1)ᵀ

**Answer:**
| Eigenvalue | Eigenvector(s) |
|---|---|
| 5 | k(1, 1, 1)ᵀ |
| 1 (twice) | k₁(−2, 1, 0)ᵀ + k₂(−1, 0, 1)ᵀ |

Since there are three independent eigenvectors, A is diagonalizable.

## Check yourself

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

1. What is the characteristic equation?
   A. lambda^3 + 7 lambda^2 + 11 lambda + 5 = 0
   B. lambda^3 - 7 lambda^2 + 5 lambda - 11 = 0
   C. lambda^3 - 11 lambda^2 + 7 lambda - 5 = 0
   D. lambda^3 - 7 lambda^2 + 11 lambda - 5 = 0
   Answer: D. trace 7, minors 11, determinant 5. (Key point: Characteristic equation)

2. The eigenvalues are
   A. 1, 1, 5
   B. 1, 5, 5
   C. 1, 2, 4
   D. -1, -1, 5
   Answer: A. (l-1)^2 (l-5). (Key point: Eigenvalues)

3. How many independent eigenvectors belong to lambda = 1?
   A. 1
   B. 2, since A - I has rank 1
   C. 3
   D. 0
   Answer: B. All rows give x + 2y + z = 0, so n - rank = 2. (Key point: Eigenvectors for 1)

4. Is A diagonalizable?
   A. No, because 1 repeats
   B. No, because it is not symmetric
   C. Yes, it has three independent eigenvectors
   D. Only if lambda = 5 repeats
   Answer: C. Geometric multiplicity equals algebraic multiplicity. (Key point: Diagonalizable?)

## Close variants set in other exams

- [Eigenvalues and eigenvectors [2,−2,2;1,1,1;…]](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aeigenvalues-and-eigenvectors-2-2-2) — Mathematics III, chapter 1, asked 8×
- [Eigenvalues and eigenvectors [1,2,−1;0,2,1;…]](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aeigenvalues-and-eigenvectors-1-2-1) — Mathematics III, chapter 1, asked 8×
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- [Eigenvalues and eigenvectors [2,1,1;−2,1,3;…]](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aeigenvalues-and-eigenvectors-2-1-1-two) — Mathematics III, chapter 1, asked 8×
- [Eigenvalues, eigenvectors, modal matrix [2,0,1;0,2,−1;…]](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aeigenvalues-eigenvectors-modal-matrix-2-0) — Mathematics III, chapter 1, asked 8×

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