Let α [ 1 2 3 4 5 6 2 1 3 5 4 6 ] and β = [ 1 2 3 4 5 6 6 1 2 4 3 5 ] . Compute each of the following. a. α − 1 b. β α c. α β

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Answer 1

Textbook Question

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Chapter 5, Problem 1E

Let α [ 1 2 3 4 5 6 2 1 3 5 4 6 ] and β = [ 1 2 3 4 5 6 6 1 2 4 3 5 ] .

Compute each of the following.

a. α − 1

b. β α

c. α β

a.

Expert Solution

To determine

To compute α−1 ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 ) .

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) then α−1=( b 1 b 2 b 3 a 1 a 2 a 3 )

Calculation:

Use the above concept to find α−1 .

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ⇒α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 ) .

Conclusion:

Thus, α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

b.

Expert Solution

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

βα=( 1 2 3 4 5 6 1 6 2 3 4 5 ) .

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

βα=( 1 2 3 4 5 6 6 1 2 4 3 5 )( 1 2 3 4 5 6 2 1 3 5 4 6 ) ∴βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

Conclusion:

Thus, βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

c.

Expert Solution

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

αβ=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ( 1 2 3 4 5 6 6 1 2 4 3 5 )=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Conclusion:

Thus, αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

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Answer 2

Textbook Question

100%

Chapter 5, Problem 1E

Let α [ 1 2 3 4 5 6 2 1 3 5 4 6 ] and β = [ 1 2 3 4 5 6 6 1 2 4 3 5 ] .

Compute each of the following.

a. α − 1

b. β α

c. α β

a.

Expert Solution

To determine

To compute α−1 ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 ) .

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) then α−1=( b 1 b 2 b 3 a 1 a 2 a 3 )

Calculation:

Use the above concept to find α−1 .

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ⇒α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 ) .

Conclusion:

Thus, α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

b.

Expert Solution

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

βα=( 1 2 3 4 5 6 1 6 2 3 4 5 ) .

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

βα=( 1 2 3 4 5 6 6 1 2 4 3 5 )( 1 2 3 4 5 6 2 1 3 5 4 6 ) ∴βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

Conclusion:

Thus, βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

c.

Expert Solution

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

αβ=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ( 1 2 3 4 5 6 6 1 2 4 3 5 )=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Conclusion:

Thus, αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

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Answer 3

Textbook Question

100%

Chapter 5, Problem 1E

Let α [ 1 2 3 4 5 6 2 1 3 5 4 6 ] and β = [ 1 2 3 4 5 6 6 1 2 4 3 5 ] .

Compute each of the following.

a. α − 1

b. β α

c. α β

a.

Expert Solution

To determine

To compute α−1 ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 ) .

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) then α−1=( b 1 b 2 b 3 a 1 a 2 a 3 )

Calculation:

Use the above concept to find α−1 .

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ⇒α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 ) .

Conclusion:

Thus, α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

b.

Expert Solution

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

βα=( 1 2 3 4 5 6 1 6 2 3 4 5 ) .

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

βα=( 1 2 3 4 5 6 6 1 2 4 3 5 )( 1 2 3 4 5 6 2 1 3 5 4 6 ) ∴βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

Conclusion:

Thus, βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

c.

Expert Solution

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

αβ=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ( 1 2 3 4 5 6 6 1 2 4 3 5 )=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Conclusion:

Thus, αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

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Answer 4

Textbook Question

100%

Chapter 5, Problem 1E

Let α [ 1 2 3 4 5 6 2 1 3 5 4 6 ] and β = [ 1 2 3 4 5 6 6 1 2 4 3 5 ] .

Compute each of the following.

a. α − 1

b. β α

c. α β

a.

Expert Solution

To determine

To compute α−1 ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 ) .

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) then α−1=( b 1 b 2 b 3 a 1 a 2 a 3 )

Calculation:

Use the above concept to find α−1 .

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ⇒α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 ) .

Conclusion:

Thus, α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

b.

Expert Solution

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

βα=( 1 2 3 4 5 6 1 6 2 3 4 5 ) .

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

βα=( 1 2 3 4 5 6 6 1 2 4 3 5 )( 1 2 3 4 5 6 2 1 3 5 4 6 ) ∴βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

Conclusion:

Thus, βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

c.

Expert Solution

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

αβ=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ( 1 2 3 4 5 6 6 1 2 4 3 5 )=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Conclusion:

Thus, αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

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Answer 5

Textbook Question

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Answer 6

Textbook Question

100%

Answer 7

a.

Expert Solution

To determine

To compute α−1 ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 ) .

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) then α−1=( b 1 b 2 b 3 a 1 a 2 a 3 )

Calculation:

Use the above concept to find α−1 .

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ⇒α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 ) .

Conclusion:

Thus, α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

b.

Expert Solution

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

βα=( 1 2 3 4 5 6 1 6 2 3 4 5 ) .

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

βα=( 1 2 3 4 5 6 6 1 2 4 3 5 )( 1 2 3 4 5 6 2 1 3 5 4 6 ) ∴βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

Conclusion:

Thus, βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

c.

Expert Solution

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

αβ=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ( 1 2 3 4 5 6 6 1 2 4 3 5 )=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Conclusion:

Thus, αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Answer 8

a.

Expert Solution

To determine

To compute α−1 ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 ) .

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) then α−1=( b 1 b 2 b 3 a 1 a 2 a 3 )

Calculation:

Use the above concept to find α−1 .

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ⇒α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 ) .

Conclusion:

Thus, α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To compute α−1 ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer 13

Answer to Problem 1E

α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

Answer 14

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 ) .

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) then α−1=( b 1 b 2 b 3 a 1 a 2 a 3 )

Calculation:

Use the above concept to find α−1 .

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ⇒α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 ) .

Conclusion:

Thus, α−1=( 1 2 3 4 5 6 2 1 3 5 4 6 )

Answer 15

b.

Expert Solution

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

βα=( 1 2 3 4 5 6 1 6 2 3 4 5 ) .

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

βα=( 1 2 3 4 5 6 6 1 2 4 3 5 )( 1 2 3 4 5 6 2 1 3 5 4 6 ) ∴βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

Conclusion:

Thus, βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer 20

Answer to Problem 1E

βα=( 1 2 3 4 5 6 1 6 2 3 4 5 ) .

Answer 21

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

βα=( 1 2 3 4 5 6 6 1 2 4 3 5 )( 1 2 3 4 5 6 2 1 3 5 4 6 ) ∴βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

Conclusion:

Thus, βα=( 1 2 3 4 5 6 1 6 2 3 4 5 )

Answer 22

c.

Expert Solution

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer to Problem 1E

αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

αβ=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ( 1 2 3 4 5 6 6 1 2 4 3 5 )=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Conclusion:

Thus, αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

To compute αβ ,where α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Answer 27

Answer to Problem 1E

αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Answer 28

Explanation of Solution

Given:

α=( 1 2 3 4 5 6 2 1 3 5 4 6 ) β=( 1 2 3 4 5 6 6 1 2 4 3 5 )

Concept Used:

If α=( a 1 a 2 a 3 b 1 b 2 b 3 ) and β=( b 1 b 2 b 3 a 1 a 2 a 3 ) then αβ=( a 1 a 2 a 3 a 1 a 2 a 3 )

Calculation:

αβ=( 1 2 3 4 5 6 2 1 3 5 4 6 ) ( 1 2 3 4 5 6 6 1 2 4 3 5 )=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Conclusion:

Thus, αβ=( 1 2 3 4 5 6 6 2 1 5 3 4 )

Answer 29

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Let α [ 1 2 3 4 5 6 2 1 3 5 4 6 ] and β = [ 1 2 3 4 5 6 6 1 2 4 3 5 ] . Compute each of the following. a. α − 1 b. β α c. α β

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Answer to Problem 1E

Explanation of Solution

Answer to Problem 1E

Explanation of Solution

Answer to Problem 1E

Explanation of Solution

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