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(

Expert Solution & Answer

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Check out a sample textbook solution

See solution

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2.7 LAB: Smallest of two numbers Instructor note: Note: this section of your textbook contains activities that you will complete for points. To ensure your work is scored, please access this page from the assignment link provided in the CTU Virtual Campus. If you did not access this page via the CTU Virtual Campus, please do so now.

I help understanding this question d'y + 4dy +3y = a, Initial Conditions: y(0) = 5 & y'(0)=0 Where a = 10 a) Find y(t) =yh(t) +yp(t) in time domainIs the system over-damped, under-damped, or critical? b) Find y(t) using Laplace Transforms

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

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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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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(