---
title: "Eigenvalues and eigenvectors [−2,2,−3;2,1,−6;…]"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:eigenvalues-and-eigenvectors-2-2-3"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/eigenvalues-and-eigenvectors-2-2-3
site: BCT Engineering
---

# Eigenvalues and eigenvectors [−2,2,−3;2,1,−6;…]

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2076 Asoj, 1+4 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-2-2-3
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

State any two properties of eigenvalues of a matrix. Obtain the eigenvalues and eigenvectors of the matrix [−2 2 −3; 2 1 −6; −1 −2 0].

## Short answer

Two properties: sum of eigenvalues equals trace, product equals determinant; A and Aᵀ share eigenvalues. The matrix has eigenvalues 5 (eigenvector (1,2,−1)ᵀ) and −3 (twice, eigenvectors (−2,1,0)ᵀ and (3,0,1)ᵀ).

## Answer

#### Two properties of eigenvalues
1. The sum of the eigenvalues of a square matrix equals its trace, and their product equals its determinant: Σλ = a₁₁ + a₂₂ + …, Πλ = |A|.
2. If λ is an eigenvalue of A, then λᵏ is an eigenvalue of Aᵏ, and (for non-singular A) 1/λ is an eigenvalue of A⁻¹. A and Aᵀ have the same eigenvalues.

(Other valid properties: the eigenvalues of a triangular matrix are its diagonal entries; eigenvalues of a Hermitian matrix are real; eigenvectors of distinct eigenvalues are independent.)

#### Eigenvalues of A = [−2 2 −3; 2 1 −6; −1 −2 0]
- Trace = −2 + 1 + 0 = −1
- Sum of principal minors = (0 − 12) + (0 − 3) + (−2 − 4) = −12 − 3 − 6 = −21
- |A| = −2(0 − 12) − 2(0 − 6) + (−3)(−4 + 1) = 24 + 12 + 9 = 45

```text
λ³ + λ² − 21λ − 45 = 0   ⇒   (λ − 5)(λ + 3)² = 0
```

**Eigenvalues: 5, −3, −3.**

#### Eigenvectors
**λ = 5:** A − 5I = [−7 2 −3; 2 −4 −6; −1 −2 −5]. From R₂: x = 2y + 3z; R₃: −x − 2y − 5z = 0 ⇒ −(2y + 3z) − 2y − 5z = 0 ⇒ y = −2z. Take z = −1: y = 2, x = 1.
X₁ = (1, 2, −1)ᵀ. Check: AX₁ = (5, 10, −5) = 5X₁ ✓.

**λ = −3:** A + 3I = [1 2 −3; 2 4 −6; −1 −2 3]. All rows give x + 2y − 3z = 0 (rank 1), so there are two independent solutions:
- y = 1, z = 0: X₂ = (−2, 1, 0)ᵀ
- y = 0, z = 1: X₃ = (3, 0, 1)ᵀ

**Answer:**
| Eigenvalue | Eigenvectors |
|---|---|
| 5 | k(1, 2, −1)ᵀ |
| −3 (twice) | k₁(−2, 1, 0)ᵀ + k₂(3, 0, 1)ᵀ |

## Check yourself

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

1. The sum of eigenvalues equals the _____ and the product equals the _____.
   A. determinant; trace
   B. trace; determinant
   C. rank; trace
   D. trace; rank
   Answer: B. Σλ = trace and Πλ = |A|. (Key point: Sum and product)

2. Trace -1, minors sum -21, |A| = 45. The eigenvalues are:
   A. 5, 3, 3
   B. -5, 3, 3
   C. 5, -3, -3
   D. -5, -3, -3
   Answer: C. λ^3 + λ^2 - 21λ - 45 = (λ-5)(λ+3)^2. (Key point: Characteristic equation)

3. For λ = -3, A + 3I reduces to x + 2y - 3z = 0 (rank 1). How many independent eigenvectors?
   A. 2
   B. 1
   C. 3
   D. 0
   Answer: A. 3 - 1 = 2 independent solutions, e.g. (-2,1,0) and (3,0,1). (Key point: Repeated eigenvalue)

4. Which is an eigenvector for λ = 5?
   A. (1, 1, 1)
   B. (1, 2, -1)
   C. (-2, 1, 0)
   D. (3, 0, 1)
   Answer: B. A(1,2,-1) = (5,10,-5) = 5(1,2,-1). (Key point: Eigenvector for 5)

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