---
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-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/eigenvalues-and-eigenvectors-2-2-3-two
site: BCT Engineering
---

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

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2072 Kartik, 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-2-2-3-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Find the eigenvalues and eigenvectors of the matrix [−2 2 −3; 2 1 −6; −1 −2 0].

## Short answer

Characteristic equation λ³+λ²−21λ−45=0 gives eigenvalues 5 and −3 (double). For λ=5, eigenvector (1,2,−1)ᵀ; for λ=−3, two independent vectors: (−2,1,0)ᵀ and (3,0,1)ᵀ.

## Answer

#### 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. Eigenvalues of the matrix:
   A. 5, 3, 3
   B. 5, −3, −3
   C. −5, 3, 3
   D. 1, 5, −3
   Answer: B. λ³+λ²−21λ−45 = (λ−5)(λ+3)². (Key point: Eigenvalues)

2. For λ = −3, the matrix A+3I has rank:
   A. 1, giving two independent eigenvectors
   B. 2, giving one eigenvector
   C. 3
   D. 0
   Answer: A. All rows are multiples of x+2y−3z = 0, so rank 1 and nullity 2. (Key point: Repeated eigenvalue)

3. Two independent eigenvectors for −3:
   A. (1,2,−1) and (3,0,1)
   B. (2,1,0) and (3,0,1)
   C. (1,0,0) and (0,1,0)
   D. (−2,1,0) and (3,0,1)
   Answer: D. Solutions of x+2y−3z=0 with (y,z) = (1,0) and (0,1). (Key point: Eigenvectors for −3)

4. How is an eigenvector verified?
   A. AX = X
   B. A + X = λ
   C. AX = λX
   D. X = Aᵀ
   Answer: C. Eigenvector satisfies AX = λX. (Key point: Check)

## Close variants set in other exams

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