Determinant Calculation and Matrix Invertibility, Assignments of Mathematics

Detailed solutions for calculating the determinants of four different matrices (a, b, c, and d) and determining the conditions for the matrices to be invertible. The solutions involve step-by-step calculations using the properties of determinants, such as row/column operations and expansion by minors. A range of matrix sizes and structures, demonstrating the application of determinant calculation techniques. It also highlights the importance of the determinant value in establishing the invertibility of a matrix, which is a crucial concept in linear algebra and its applications in various fields, including mathematics, physics, engineering, and computer science. The comprehensive nature of the solutions presented in this document makes it a valuable resource for students and professionals seeking to deepen their understanding of determinant calculations and matrix invertibility.

Typology: Assignments

2022/2023

Uploaded on 10/26/2023

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I.2.1. Tính giá tr các định thc
a) A = 512 16 17
2 5 6 6
3 7 10 10
1 2 3 3;
Bài gii
BĐSC, ta được
A = 512 16 17
2 5 6 6
3 7 10 10
1 2 3 3 14
dd
→
1 2 3 3
2 5 6 6
3 7 10 10
5 12 16 17
221
331
441
2
3
5
ddd
ddd
ddd
→−
→−
→−
→
1 2 3 3
0 1 0 0
0 1 1 1
0 2 1 2
Khai trin theo ct đầu, ta được
A = -1.(-1)1+11 0 0
1 1 1
2 1 2 = (1).(1.1.2+1.1.0+0.1.20.1.21.1.11.0.2)
= 1
Vy detA = 1
b) B = 616 13 20
2 5 4 6
3 8 7 9
1 3 2 4;
Bài gii
BĐSC, ta được
B = 616 13 20
2 5 4 6
3 8 7 9
1 3 2 4 14
dd
→
1 3 2 4
2 5 4 6
3 8 7 9
616 13 20
1
221
33
441
2
3
6
ddd
ddd
ddd
→−
→−
→−
→
1324
01 0 2
01 1 3
02 1 4
Khai trin theo ct đầu, ta được
B = -1.(-1)1+11 0 2
1 1 3
2 1 4
pf3

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I.2.1. Tính giá trị các định thức

a) A =

Bài giải

BĐSC, ta được

A =

 d^1^^ ↔ d^4 → −

2 2 1 3 3 1 4 4 1

2 3 5

d d d d d d d d d

→ − → −

Khai triển theo cột đầu, ta được

A = -1.(-1)1+1

Vậy detA = − 1

b) B =

Bài giải

BĐSC, ta được

B =

(^)  d^1^^ ↔ d^4 → −

1

2 2 1 3 3 4 4 1

2 3 6

d d d d d d d d d

→ − → −

Khai triển theo cột đầu, ta được

B = -1.(-1)1+1

= (−1). [(−1).1. (−4) + (−1).1. (−2) + 0. (−3). (−2) − (−2).1. (−2) −

Vậy detB = 1

c) C =

Bài giải

C =

2 2 1 3 3 1 4 4 1

d d d d d d d d d

→ − → − →→ −

3 3 2 4 4 2

2 3

d d d d d d

→ −  → − →

(^)  d^4^^ → d^^^4 −^3 d^3 →

Vậy detC = 12

d) D =

Khai triển theo hàng 3, ta được

Vậy detD = 0

I.2.2. Tìm điều kiện của tham số thực m để ma trận đã cho khả nghịch

a) A = 2

m m m

       +  Bài giải

detA = 2

m m + m

= m^3^ − 9 m 2 − 25 m − 225

(A khả nghịch) ⇔ (det A = m^3^ − 9 m^2 − 25 m − 225 ≠ 0) ⇔ ( m ∉ −{ 5,5,9 )}.