---
title: "Eigenvalues and eigenvectors [3,−4,4;1,−2,4;…]"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:eigenvalues-and-eigenvectors-3-4-4"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/eigenvalues-and-eigenvectors-3-4-4
site: BCT Engineering
---

# Eigenvalues and eigenvectors [3,−4,4;1,−2,4;…]

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2068 Chaitra, 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-3-4-4
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Find the eigenvalues and eigenvectors of A = [3 −4 4; 1 −2 4; 1 −1 3].

## Short answer

Eigenvalues: −1, 2, 3. Egenvectors: for −1, k(1,1,0)^T; for 2, k(0,1,1)^T; for 3, k(1,1,1)^T.

## Answer

A = [3 −4 4; 1 −2 4; 1 −1 3]

#### Characteristic equation
- Trace = 3 − 2 + 3 = 4
- Sum of principal minors = (−6 + 4) + (9 − 4) + (−6 + 4) = −2 + 5 − 2 = 1
- |A| = 3(−6 + 4) + 4(3 − 4) + 4(−1 + 2) = −6 − 4 + 4 = −6

```text
λ³ − 4λ² + λ + 6 = 0
(λ + 1)(λ − 2)(λ − 3) = 0
```

**Eigenvalues: λ = −1, 2, 3.**

#### Eigenvectors
**λ = −1:** A + I = [4 −4 4; 1 −1 4; 1 −1 4]. Equations x − y + z = 0 and x − y + 4z = 0 ⇒ z = 0, x = y.
X₁ = (1, 1, 0)ᵀ.

**λ = 2:** A − 2I = [1 −4 4; 1 −4 4; 1 −1 1]. Equations x − 4y + 4z = 0 and x − y + z = 0. Subtract: −3y + 3z = 0 ⇒ y = z, and then x = 0.
X₂ = (0, 1, 1)ᵀ.

**λ = 3:** A − 3I = [0 −4 4; 1 −5 4; 1 −1 0]. R₁: y = z; R₃: x = y.
X₃ = (1, 1, 1)ᵀ.

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

(Check: sum of eigenvalues −1 + 2 + 3 = 4 = trace; product −6 = |A|.)

## Check yourself

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

1. Which gives the coefficients of λ³ − (trace)λ² + (sum of principal minors)λ − |A|?
   A. Trace 4, minors sum 5, |A| = 6
   B. Trace 4, minors sum 1, |A| = −6
   C. Trace 2, minors sum 1, |A| = −6
   D. Trace 4, minors sum 1, |A| = 6
   Answer: B. Trace = 4, principal minors sum = 1, |A| = −6, giving λ³ − 4λ² + λ + 6 = 0. (Key point: Characteristic equation)

2. What are the eigenvalues?
   A. −1, 2, 3
   B. 1, 2, 3
   C. −1, −2, 3
   D. −1, 2, −3
   Answer: A. λ³ − 4λ² + λ + 6 factors as (λ+1)(λ−2)(λ−3). (Key point: Eigenvalues)

3. How is the eigenvector for a given λ found?
   A. Solve AX = B
   B. Compute A⁻¹λ
   C. Solve (A + λI)X = I
   D. Solve (A − λI)X = 0 for non-zero X
   Answer: D. Eigenvectors are non-zero solutions of (A − λI)X = 0. (Key point: Eigenvector method)

4. For λ = −1, A + I gives z = 0 and x = y. The eigenvector is:
   A. (1, 0, 1)
   B. (0, 1, 1)
   C. (1, 1, 0)
   D. (1, 1, 1)
   Answer: C. z = 0, x = y gives (1, 1, 0). (Key point: Eigenvector for λ = −1)

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