---
title: "Modal matrix of a matrix [5,4;1,2]"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:modal-matrix-of-a-matrix-5"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/modal-matrix-of-a-matrix-5
site: BCT Engineering
---

# Modal matrix of a matrix [5,4;1,2]

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2078 Kartik, 5 marks; 2075 Asoj, 5 marks (asked 2 times)
- Video: [Find the model matrix of [[4,1],[2,3]] and verify its diagonalization. ](https://www.youtube.com/watch?v=C8CvogeYJOQ) — SK-Academy Of Mathematics 
- 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-5
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

For the matrix A = [5 4; 1 2], find the modal matrix and the corresponding diagonal matrix.

## Short answer

Eigenvalues are 1 and 6. Corresponding eigenvectors are (1, -1) and (4, 1). Modal matrix P = [1 4; -1 1] and diagonal matrix D = diag(1, 6).

## Answer

The modal matrix P has the eigenvectors of A as its columns, and P⁻¹AP = D is diagonal.

A = [5 4; 1 2]

#### Eigenvalues
```text
|A − λI| = (5 − λ)(2 − λ) − 4 = λ² − 7λ + 6 = (λ − 1)(λ − 6) = 0
```
**λ = 1, 6.**

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

**λ = 6:** (A − 6I)X = [−1 4; 1 −4]X = 0 ⇒ x = 4y. X₂ = (4, 1)ᵀ.

#### Modal matrix and diagonal matrix
```text
P = [ 1  4]        |P| = 1 + 4 = 5 ≠ 0
    [−1  1]

P⁻¹ = (1/5) [1 −4]
            [1  1]
```

Verification:

```text
AP = [5−4   20+4]  = [1 24]
     [1−2    4+2]    [−1 6]

P⁻¹AP = (1/5)[1 −4; 1 1][1 24; −1 6] = (1/5)[5 0; 0 30] = [1 0; 0 6]
```

**Answer: P = [1 4; −1 1] and D = P⁻¹AP = diag(1, 6).**

## Check yourself

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

1. Which equation gives the eigenvalues?
   A. lambda^2 + 7 lambda + 6 = 0
   B. lambda^2 - 7 lambda + 6 = 0
   C. lambda^2 - 7 lambda - 6 = 0
   D. lambda^2 - 6 lambda + 7 = 0
   Answer: B. Trace 7, determinant 10 - 4 = 6, giving roots 1 and 6. (Key point: Eigenvalues)

2. The eigenvector for lambda = 6 is
   A. (4, 1)
   B. (1, 4)
   C. (1, -1)
   D. (4, -1)
   Answer: A. -x + 4y = 0 gives x = 4y. (Key point: Eigenvector for 6)

3. P^-1 A P equals
   A. diag(6, 1)
   B. diag(5, 2)
   C. diag(1, 6)
   D. [1 4; -1 1]
   Answer: C. With columns ordered X1 (lambda 1) and X2 (lambda 6). (Key point: Diagonal matrix)

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