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linear algebra and matrix theory, Exercises of Agricultural Mathematics

Exercises about linear algebra and matrix theory

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2021/2022

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MATH 122 โ€“ H Activity 2
Linear Algebra and Matrix Theory Regine Joyce A. Camacho โ€“ BS Mathematics
I. MODIFIED TRUE OR FALSE.
1. FALSE. Type 3
2. FALSE. Type 2
3. FALSE. Only trivial solution
4. TRUE
5. FALSE. Has no zero row.
II. COMPUTATIONS.
1. Let ๐ด=[2 โˆ’1
3 2] and ๐ต=[โˆ’4 2
1 3]. Using these matrices, show that
a. (๐ด+๐ต)2โ‰ ๐ด2+2๐ด๐ต+๐ต2
Solution:
Given that ๐ด=[2 โˆ’1
3 2] and ๐ต=[โˆ’4 2
1 3], ๐ด+๐ต is
[2 โˆ’1
3 2]+[โˆ’4 2
1 3]=[โˆ’2 1
4 5]
With that, (๐ด+๐ต)2=(๐ด+๐ต)(๐ด+๐ต) is
[โˆ’2 1
4 5][โˆ’2 1
4 5]=[(โˆ’2)(โˆ’2)+(1)(4) (โˆ’2)(1)+(1)(5)
(4)(โˆ’2)+(5)(4) (4)(1)+(5)(5)]
=[4+4 โˆ’2+5
โˆ’8+20 4+25]
โ‡’(๐ด+๐ต)2=[8 3
12 29]
Since ๐ด=[2 โˆ’1
3 2], this implies that ๐ด2=(๐ด)(๐ด) is
[2 โˆ’1
3 2][2 โˆ’1
3 2]=[(2)(2)+(โˆ’1)(3) (2)(โˆ’1)+(โˆ’1)(2)
(3)(2)+(2)(3) (3)(โˆ’1)+(2)(2)]
=[4โˆ’3 โˆ’2โˆ’2
6+6 โˆ’3+4]
โ‡’๐ด2=[1 โˆ’4
12 1]
Since ๐ด=[2 โˆ’1
3 2] and ๐ต=[โˆ’4 2
1 3], this implies that ๐ด๐ต is
[2 โˆ’1
3 2][โˆ’4 2
1 3]=[(2)(โˆ’4)+(โˆ’1)(1) (2)(2)+(โˆ’1)(3)
(3)(โˆ’4)+(2)(1) (3)(2)+(2)(3)]
=[โˆ’8โˆ’1 4โˆ’3
โˆ’12+2 6+6]
โ‡’๐ด๐ต=[โˆ’9 1
โˆ’10 12]
With that, 2๐ด๐ต is 2[โˆ’9 1
โˆ’10 12]=[โˆ’18 2
โˆ’20 24]
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MATH 122 โ€“ H Activity 2

Linear Algebra and Matrix Theory Regine Joyce A. Camacho โ€“ BS Mathematics

I. MODIFIED TRUE OR FALSE.

  1. FALSE. Type 3
  2. FALSE. Type 2
  3. FALSE. Only trivial solution

4. TRUE

  1. FALSE. Has no zero row.

II. COMPUTATIONS.

  1. Let ๐ด = [

] and ๐ต = [

]. Using these matrices, show that

a. (๐ด + ๐ต)

2

2

2

Solution:

Given that ๐ด = [

]

and ๐ต = [

]

, ๐ด + ๐ต is

[

]

[

]

[

]

With that, (๐ด + ๐ต)

2

= (๐ด + ๐ต)(๐ด + ๐ต) is

[

] [

]

[

]

[

]

2

= [

]

Since ๐ด = [

], this implies that ๐ด

2

is

[

] [

] = [

]

[

]

2

= [

]

Since ๐ด = [

] and ๐ต = [

], this implies that ๐ด๐ต is

[

] [

] = [

]

= [

]

โ‡’ ๐ด๐ต = [

]

With that, 2 ๐ด๐ต is

2 [

] = [

]

Since ๐ต = [

], this implies that ๐ต

2

= (๐ต)(๐ต) is

[

] [

] = [

]

= [

]

2

= [

]

So, ๐ด

2

2

is

[

] + [

] + [

] = [

]

Moreover, [

] โ‰  [

]. Therefore, this shows (๐ด + ๐ต)

2

2

2

b. (๐ด + ๐ต)(๐ด โˆ’ ๐ต) โ‰  ๐ด

2

2

We obtain from the previous item that ๐ด + ๐ต = [

], ๐ด

2

= [

], and ๐ต

2

[

].

Given that ๐ด = [

] and ๐ต = [

], ๐ด โˆ’ ๐ต is

[

] โˆ’ [

] = [

]

So,

is

[

] [

] = [

]

[

]

= [

]

Since ๐ด

2

= [

] and ๐ต

2

= [

], then ๐ด

2

2

is

[

] โˆ’ [

] = [

]

Moreover, [

] โ‰  [

]. Therefore, this shows that (๐ด + ๐ต)(๐ด โˆ’ ๐ต) โ‰ 

2

2

  1. If ๐ด = [

] and ๐ต = [

] and ๐ด๐‘‹ = ๐ต, then the matrix ๐‘‹ is _______.

We substitute the value of ๐‘ฅ 12

to 4 ๐‘ฅ

12

22

= 2 , so

22

22

22

22

22

Since we have now the value of ๐‘ฅ

22

, we can now substitute it to 4 ๐‘ฅ

12

22

= 2 , hence

12

12

12

12

With that, [

11

12

21

22

] = [

]. Therefore, the matrix ๐‘‹ = [

].

3. [

]

Solution:

[

]

[

] ๐‘…

2

1

[

]

3

2

[

] โˆ’ 2 ๐‘…

3

2

2

[

] โˆ’ 2 ๐‘…

3

2

2