a. Proof: Consider a 1 × 1 matrix, A = [ a ] Then, det ( A ) = a Now, we have − A = [ − a ] Then, we get det ( − A ) = − a = − det ( A ) Similarly consider a 3 × 3 matrix as given below: A = [ a b c 0 d e 0 f

Question

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Answer 1

Expert Solution & Answer

Chapter 2.6, Problem 4P

Explanation of Solution

a.

Proof:

Consider a 1×1 matrix,

A=[a]

Then, det(A)=a

Now, we have

−A=[−a]

Then, we get

det(−A)=−a =−det(A)

Similarly consider a 3×3 matrix as given below:

A=[a b c0 d e0 f

Explanation of Solution

b.

Proof:

Let us consider a 2×2 matrix as given below:

A=[a     bc     d]

Then, we get det(A)=ad−bc

Now, we have

−A=[−a    -b−c    -d]

Then, we get

det(−A)=ad−bc             =det(A)

Similarly consider a 4×4 matrix as,

A=[a     b     c      d0     e      f      g0     h      0    �

Explanation of Solution

Generalize the results of part (a.) and (b.):

From part a.) and part b.), we conclude that

For any n×n matrix,

det(−A)=−det(

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Check out a sample textbook solution

See solution

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Ideal MOSFET Current–Voltage Characteristics—NMOS Device and draw the circuit

1. Create a Person.java file. Implement the public Person and Student classes in Person.java, including all the variables and methods in the UMLS. Person -name: String -street: String -city: String +Person(String name, String, street, String, city) +getName(): String +setName(String name): void +getStreet(): String +setStreet(String street): void +getCity(): String +setCity(String City): void +toString(): String Student -Id: int +Person(String name, String, street, String, city, int Id) +getId(): int +setId(int Id): void +toString(): String 2. Create a StudentTest.java file. Implement a public StudentTest class with a main method. In the main method, create one student object and print the object using System.out.println(). Your printing result must follow the example output: name: Mike, street: Morris Ave, city: Union, Id: 1000 Hint: You need to modify the toString methods in the Student class and Person class!

1) Apply the Paint Blue algorithm discussed in class to the following Finite Automata. a a a b b a COIS-3050H-R-W01-2025WI-COMB Formal Languages & Automata a b Show the status of the Finite Automata at the conclusion of the Paint Blue Algorithm (mark the visited states with an X and only include edges that have not been followed). 2) Use the pumping lemma to prove the following language is nonregular: L= {ab} = {abbb, aabbbbbb, aaabbbbbbbbb, ...}

Answer 2

Expert Solution & Answer

Chapter 2.6, Problem 4P

Explanation of Solution

a.

Proof:

Consider a 1×1 matrix,

A=[a]

Then, det(A)=a

Now, we have

−A=[−a]

Then, we get

det(−A)=−a =−det(A)

Similarly consider a 3×3 matrix as given below:

A=[a b c0 d e0 f

Explanation of Solution

b.

Proof:

Let us consider a 2×2 matrix as given below:

A=[a     bc     d]

Then, we get det(A)=ad−bc

Now, we have

−A=[−a    -b−c    -d]

Then, we get

det(−A)=ad−bc             =det(A)

Similarly consider a 4×4 matrix as,

A=[a     b     c      d0     e      f      g0     h      0    �

Explanation of Solution

Generalize the results of part (a.) and (b.):

From part a.) and part b.), we conclude that

For any n×n matrix,

det(−A)=−det(

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

Chapter 2.6, Problem 4P

Explanation of Solution

a.

Proof:

Consider a 1×1 matrix,

A=[a]

Then, det(A)=a

Now, we have

−A=[−a]

Then, we get

det(−A)=−a =−det(A)

Similarly consider a 3×3 matrix as given below:

A=[a b c0 d e0 f

Explanation of Solution

b.

Proof:

Let us consider a 2×2 matrix as given below:

A=[a     bc     d]

Then, we get det(A)=ad−bc

Now, we have

−A=[−a    -b−c    -d]

Then, we get

det(−A)=ad−bc             =det(A)

Similarly consider a 4×4 matrix as,

A=[a     b     c      d0     e      f      g0     h      0    �

Explanation of Solution

Generalize the results of part (a.) and (b.):

From part a.) and part b.), we conclude that

For any n×n matrix,

det(−A)=−det(

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 2.6, Problem 4P

Explanation of Solution

a.

Proof:

Consider a 1×1 matrix,

A=[a]

Then, det(A)=a

Now, we have

−A=[−a]

Then, we get

det(−A)=−a =−det(A)

Similarly consider a 3×3 matrix as given below:

A=[a b c0 d e0 f

Explanation of Solution

b.

Proof:

Let us consider a 2×2 matrix as given below:

A=[a     bc     d]

Then, we get det(A)=ad−bc

Now, we have

−A=[−a    -b−c    -d]

Then, we get

det(−A)=ad−bc             =det(A)

Similarly consider a 4×4 matrix as,

A=[a     b     c      d0     e      f      g0     h      0    �

Explanation of Solution

Generalize the results of part (a.) and (b.):

From part a.) and part b.), we conclude that

For any n×n matrix,

det(−A)=−det(

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 2.6, Problem 4P

Explanation of Solution

a.

Proof:

Consider a 1×1 matrix,

A=[a]

Then, det(A)=a

Now, we have

−A=[−a]

Then, we get

det(−A)=−a =−det(A)

Similarly consider a 3×3 matrix as given below:

A=[a b c0 d e0 f

Explanation of Solution

b.

Proof:

Let us consider a 2×2 matrix as given below:

A=[a     bc     d]

Then, we get det(A)=ad−bc

Now, we have

−A=[−a    -b−c    -d]

Then, we get

det(−A)=ad−bc             =det(A)

Similarly consider a 4×4 matrix as,

A=[a     b     c      d0     e      f      g0     h      0    �

Explanation of Solution

Generalize the results of part (a.) and (b.):

From part a.) and part b.), we conclude that

For any n×n matrix,

det(−A)=−det(

Answer 8

Chapter 2.6, Problem 4P

Explanation of Solution

a.

Proof:

Consider a 1×1 matrix,

A=[a]

Then, det(A)=a

Now, we have

−A=[−a]

Then, we get

det(−A)=−a =−det(A)

Similarly consider a 3×3 matrix as given below:

A=[a b c0 d e0 f

Answer 9

Chapter 2.6, Problem 4P

Explanation of Solution

a.

Proof:

Consider a 1×1 matrix,

A=[a]

Then, det(A)=a

Now, we have

−A=[−a]

Then, we get

det(−A)=−a =−det(A)

Similarly consider a 3×3 matrix as given below:

A=[a b c0 d e0 f

Answer 10

Explanation of Solution

b.

Proof:

Let us consider a 2×2 matrix as given below:

A=[a     bc     d]

Then, we get det(A)=ad−bc

Now, we have

−A=[−a    -b−c    -d]

Then, we get

det(−A)=ad−bc             =det(A)

Similarly consider a 4×4 matrix as,

A=[a     b     c      d0     e      f      g0     h      0    �

Answer 11

Explanation of Solution

b.

Proof:

Let us consider a 2×2 matrix as given below:

A=[a     bc     d]

Then, we get det(A)=ad−bc

Now, we have

−A=[−a    -b−c    -d]

Then, we get

det(−A)=ad−bc             =det(A)

Similarly consider a 4×4 matrix as,

A=[a     b     c      d0     e      f      g0     h      0    �

Answer 12

Explanation of Solution

Generalize the results of part (a.) and (b.):

From part a.) and part b.), we conclude that

For any n×n matrix,

det(−A)=−det(

Answer 13

Explanation of Solution

Generalize the results of part (a.) and (b.):

From part a.) and part b.), we conclude that

For any n×n matrix,

det(−A)=−det(

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

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Check out a sample textbook solution

See solution

Answer 17

Ch. 2.1 - Prob. 1P Ch. 2.1 - Prob. 2P Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - Prob. 7P Ch. 2.2 - Prob. 1P Ch. 2.3 - Prob. 1P Ch. 2.3 - Prob. 2P

a. Proof: Consider a 1 × 1 matrix, A = [ a ] Then, det ( A ) = a Now, we have − A = [ − a ] Then, we get det ( − A ) = − a = − det ( A ) Similarly consider a 3 × 3 matrix as given below: A = [ a b c 0 d e 0 f

Solution Summary: The author illustrates how a 1times 1 matrix, A=left[aright], and mathrmdet(A)=ad-bc

Explanation of Solution

a.

Proof:

Consider a 1×1 matrix,

A=[a]

Then, det(A)=a

Now, we have

−A=[−a]

Then, we get

det(−A)=−a =−det(A)

Similarly consider a 3×3 matrix as given below:

A=[a b c0 d e0 f

Explanation of Solution

Explanation of Solution

Generalize the results of part (a.) and (b.):

From part a.) and part b.), we conclude that

For any n×n matrix,

det(−A)=−det(

Want to see the full answer?

Chapter 2 Solutions

a. Proof: Consider a 1 × 1 matrix, A = [ a ] Then, det ( A ) = a Now, we have − A = [ − a ] Then, we get det ( − A ) = − a = − det ( A ) Similarly consider a 3 × 3 matrix as given below: A = [ a b c 0 d e 0 f

Solution Summary: The author illustrates how a 1times 1 matrix, A=left[aright], and mathrmdet(A)=ad-bc

Explanation of Solution

a. Proof: Consider a 1×1 matrix, A=[a] Then, det(A)=a Now, we have −A=[−a] Then, we get det(−A)=−a =−det(A) Similarly consider a 3×3 matrix as given below: A=[a b c0 d e0 f

Explanation of Solution

b. Proof: Let us consider a 2×2 matrix as given below: A=[a bc d] Then, we get det(A)=ad−bc Now, we have −A=[−a -b−c -d] Then, we get det(−A)=ad−bc =det(A) Similarly consider a 4×4 matrix as, A=[a b c d0 e f g0 h 0 �

Explanation of Solution

Generalize the results of part (a.) and (b.): From part a.) and part b.), we conclude that For any n×n matrix, det(−A)=−det(

Want to see the full answer?

Chapter 2 Solutions

a.

Proof:

Consider a 1×1 matrix,

A=[a]

Then, det(A)=a

Now, we have

−A=[−a]

Then, we get

det(−A)=−a =−det(A)

Similarly consider a 3×3 matrix as given below:

A=[a b c0 d e0 f

b.

Proof:

Let us consider a 2×2 matrix as given below:

A=[a bc d]

Then, we get det(A)=ad−bc

Now, we have

−A=[−a -b−c -d]

Then, we get

det(−A)=ad−bc =det(A)

Similarly consider a 4×4 matrix as,

A=[a b c d0 e f g0 h 0 �

Generalize the results of part (a.) and (b.):

From part a.) and part b.), we conclude that

For any n×n matrix,

det(−A)=−det(