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

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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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make corrections of this program based on the errors shown. this is CIS 227 .

Create 6 users: Don, Liz, Shamir, Jose, Kate, and Sal. Create 2 groups: marketing and research. Add Shamir, Jose, and Kate to the marketing group. Add Don, Liz, and Sal to the research group. Create a shared directory for each group. Create two files to put into each directory: spreadsheetJanuary.txt meetingNotes.txt Assign access permissions to the directories: Groups should have Read+Write access Leave owner permissions as they are “Everyone else” should not have any access Submit for grade: Screenshot of /etc/passwd contents showing your new users Screenshot of /etc/group contents showing new groups with their members Screenshot of shared directories you created with files and permissions

⚫ your circuit diagrams for your basic bricks, such as AND, OR, XOR gates and 1 bit multiplexers, ⚫ your circuit diagrams for your extended full adder, designed in Section 1 and ⚫ your circuit diagrams for your 8-bit arithmetical-logical unit, designed in Section 2. 1 An Extended Full Adder In this Section, we are going to design an extended full adder circuit (EFA). That EFA takes 6 one bit inputs: aj, bj, Cin, Tin, t₁ and to. Depending on the four possible combinations of values on t₁ and to, the EFA produces 3 one bit outputs: sj, Cout and rout. The EFA can be specified in principle by a truth table with 26 = 64 entries and 3 outputs. However, as the EFA ignores certain inputs in certain cases, it is easier to work with the following overview specification, depending only on t₁ and to in the first place: t₁ to Description 00 Output Relationship Ignored Inputs Addition Mode 2 Coutsjaj + bj + Cin, Tout= 0 Tin 0 1 Shift Left Mode Sj = Cin, Cout=bj, rout = 0 rin, aj 10 1 1 Shift Right…

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

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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(