---
title: "Eigenvalues and eigenvectors [1,2,−1;0,2,1;…]"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:eigenvalues-and-eigenvectors-1-2-1"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/eigenvalues-and-eigenvectors-1-2-1
site: BCT Engineering
---

# Eigenvalues and eigenvectors [1,2,−1;0,2,1;…]

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2081 Bhadra, 2+3 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-1-2-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Find the eigenvalues and eigenvectors of the matrix [1 2 −1; 0 2 1; 0 0 3].

## Short answer

The matrix is upper triangular, so eigenvalues are the diagonal entries: 1, 2,3. Solving (A − λI)X = 0 yields eigenvectors: for λ = 1, k(1,0,0)ᵀ; for λ = 2, k(2,1,0)ᵀ; for λ = 3, k(1,2,2)ᵀ.

## Answer

A = [1 2 −1; 0 2 1; 0 0 3]

#### Eigenvalues
A is upper triangular, so the eigenvalues are the diagonal entries:

```text
|A − λI| = (1 − λ)(2 − λ)(3 − λ) = 0
```
**λ = 1, 2, 3.**

#### Eigenvectors
**λ = 1:** A − I = [0 2 −1; 0 1 1; 0 0 2]. This gives z = 0, y = 0, x free.
X₁ = (1, 0, 0)ᵀ.

**λ = 2:** A − 2I = [−1 2 −1; 0 0 1; 0 0 1]. This gives z = 0, x = 2y.
X₂ = (2, 1, 0)ᵀ.

**λ = 3:** A − 3I = [−2 2 −1; 0 −1 1; 0 0 0]. This gives y = z and −2x + 2y − z = 0 ⇒ x = y/2. Take y = 2:
X₃ = (1, 2, 2)ᵀ.

**Answer:**
| Eigenvalue | Eigenvector |
|---|---|
| 1 | k(1, 0, 0)ᵀ |
| 2 | k(2, 1, 0)ᵀ |
| 3 | k(1, 2, 2)ᵀ |

The three eigenvalues are distinct, so the eigenvectors are linearly independent.

## Check yourself

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

1. The eigenvalues are
   A. 0, 1, 2
   B. 1, 2, 3 (the diagonal entries)
   C. 1, 1, 3
   D. -1, -2, -3
   Answer: B. The matrix is upper triangular. (Key point: Eigenvalues)

2. For lambda = 3 which vector is correct?
   A. (1, 2, 2)
   B. (1, 0, 0)
   C. (2, 1, 0)
   D. (1, 1, 1)
   Answer: A. y = z and x = y/2, with y = 2. (Key point: Eigenvector for 3)

3. Why are the eigenvectors linearly independent?
   A. The matrix is symmetric
   B. The determinant is zero
   C. The three eigenvalues are distinct
   D. They are all unit vectors
   Answer: C. Eigenvectors of distinct eigenvalues are independent. (Key point: Independence)

## Close variants set in other exams

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