---
title: "Modal matrix of a matrix [2,1,1;−2,1,3;…]"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:modal-matrix-of-a-matrix-2"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/modal-matrix-of-a-matrix-2
site: BCT Engineering
---

# Modal matrix of a matrix [2,1,1;−2,1,3;…]

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2074 Chaitra, 5 marks
- Video: [Modal Matrix and Diagonalising a Given Matrix](https://www.youtube.com/watch?v=FrtBbJRNipI) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/modal-matrix-of-a-matrix-2
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Find the modal matrix for the matrix A = [2 1 1; −2 1 3; 2 1 −1].

## Short answer

Eigenvalues: −2, 2, 2. Only one independent eigenvector for λ=2, so matrix is defective. Modal matrix P = [0 4 1; −1 −2 3; 1 2 1] (columns: eigenvector (0,−1,1), 2×(2,−1,1), generalized (1,3,1)). P⁻¹AP gives Jordan form diag(−2,2,2) with a 1 in the (2,3) entry.

## Answer

A = [2 1 1; −2 1 3; 2 1 −1]

#### Eigenvalues
- 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
```
**λ = −2, 2, 2.**

#### Eigenvectors
**λ = −2:** A + 2I = [4 1 1; −2 3 3; 2 1 1]. From R₁ and R₃: 2x = 0 ⇒ x = 0, and 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 ⇒ x = 2z. Rank = 2, so only one independent vector:
X₂ = (2, −1, 1)ᵀ.

#### The modal matrix
The repeated eigenvalue 2 has only one independent eigenvector, so A is **defective** and cannot be reduced to a diagonal matrix. The modal matrix is then built with a **generalised eigenvector** X₃ satisfying (A − 2I)X₃ = (an eigenvector for λ = 2).

Use the eigenvector 2X₂ = (4, −2, 2)ᵀ. Solve (A − 2I)X₃ = (4, −2, 2)ᵀ: y + z = 4 and −2x − y + 3z = −2. Taking z = 1 gives y = 3, x = 1:

X₃ = (1, 3, 1)ᵀ. Check: (A − 2I)X₃ = (4, −2, 2)ᵀ ✓.

```text
P = [ 0  4  1]
    [−1 −2  3]        |P| = 16 ≠ 0
    [ 1  2  1]
```
(columns: X₁, 2X₂, X₃). Then

```text
P⁻¹AP = J = [−2  0  0]
            [ 0  2  1]
            [ 0  0  2]
```

**Answer: the modal matrix is P above; it gives the Jordan form J, not a purely diagonal form, because A has only two independent eigenvectors.**

## Check yourself

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

1. Which are the eigenvalues of A?
   A. 2, 2, 2
   B. -2, 2, 2
   C. -2, -2, 2
   D. -2, 1, 2
   Answer: B. λ^3 - 2λ^2 - 4λ + 8 = (λ-2)^2(λ+2). (Key point: Eigenvalues)

2. Why can A not be diagonalized?
   A. λ = 2 is repeated but has only one independent eigenvector
   B. It has a zero eigenvalue
   C. |A| = 0
   D. Eigenvalues are complex
   Answer: A. A - 2I has rank 2, giving one eigenvector, so A is defective. (Key point: Defective matrix)

3. How is X3 defined?
   A. (A + 2I)X3 = 0
   B. (A - 2I)X3 = an eigenvector for λ = 2
   C. A X3 = X3
   D. X3 = X1 + X2
   Answer: B. (A - 2I)X3 = (4,-2,2)^T gives X3 = (1,3,1)^T. (Key point: Generalised eigenvector)

4. P^-1AP is:
   A. Diagonal diag(-2, 2, 2)
   B. The identity
   C. Upper triangular with all zero
   D. Jordan form with a 1 above the repeated 2
   Answer: D. The result is J with entries -2, 2, 2 and a single 1 linking the repeated eigenvalue. (Key point: Resulting form)

## Related in Mathematics III

- [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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