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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USING PYTHON A tridiagonal matrix is one where the only nonzero elements are the ones on the main diagonal (i.e., ai,j where j = i) and the ones immediately above and belowit(i.e.,ai,j wherej=i+1orj=i−1). Write a function that solves a linear system whose coefficient matrix is tridiag- onal. In this case, Gauss elimination can be made much more efficient because most elements are already zero and don’t need to be modified or added. Please show steps and explain.

Consider the following. 0 1 -3 A = 0 4 0 4 1 2 2 2 -1 2 (a) Verify that A is diagonalizable by computing P-AP. p-1AP = (b) Use the result of part (a) and the theorem below to find the eigenvalues of A. Similar Matrices Have the Same Eigenvalues If A and B are similar n x n matrices, then they have the same eigenvalues.

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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(