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

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

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2071 Shrawan, 5 marks
- 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-2-1-1-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

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

## Short answer

The characteristic equation λ³−2λ²−4λ+8=0 yields λ=2 (double) and −2. For λ=−2, eigenvector (0,−1,1)ᵀ; for λ=2, the only independent vector is (2,−1,1)ᵀ, so the matrix is not diagonalizable.

## Answer

A = [2 1 1; −2 1 3; 2 1 −1]

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

```text
λ³ − 2λ² − 4λ + 8 = 0   ⇒   (λ − 2)²(λ + 2) = 0
```

**Eigenvalues: λ = 2, 2, −2.**

#### Eigenvectors
**λ = −2:** A + 2I = [4 1 1; −2 3 3; 2 1 1]. R₁ − R₃ gives 2x = 0 ⇒ x = 0, and then y + z = 0.
X₁ = (0, −1, 1)ᵀ.

**λ = 2:** A − 2I = [0 1 1; −2 −1 3; 2 1 −3]. R₁: y = −z. R₂: −2x − y + 3z = 0 ⇒ −2x + z + 3z = 0 ⇒ x = 2z. R₃ is then satisfied. Rank = 2, so there is only one independent vector (take z = 1):
X₂ = (2, −1, 1)ᵀ.

**Answer:**
| Eigenvalue | Eigenvector |
|---|---|
| −2 | k(0, −1, 1)ᵀ |
| 2 (twice) | k(2, −1, 1)ᵀ |

The repeated eigenvalue 2 has only one independent eigenvector, so A is not diagonalizable.

## Check yourself

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

1. The eigenvalues:
   A. −2, −2, 2
   B. 2, 2, −2
   C. 2, 1, −1
   D. 4, −2, 1
   Answer: B. λ³−2λ²−4λ+8 = (λ−2)²(λ+2). (Key point: Eigenvalues)

2. For repeated λ=2, how many independent eigenvectors?
   A. One
   B. Two
   C. Three
   D. Zero
   Answer: A. A−2I has rank 2, so nullity 1. (Key point: Independent eigenvectors)

3. Eigenvector for −2:
   A. (2,−1,1)
   B. (1,1,0)
   C. (0,1,1)
   D. (0,−1,1)
   Answer: D. x = 0 and y+z = 0. (Key point: Eigenvector λ=−2)

4. Is A diagonalizable?
   A. Yes, always
   B. Yes, since eigenvalues are real
   C. No, repeated eigenvalue 2 has fewer independent eigenvectors than multiplicity
   D. No, since |A| < 0
   Answer: C. Only two independent eigenvectors exist in total. (Key point: Diagonalizability)

## Close variants set in other exams

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