Zookal
Zookal

We’d love to hear any feedback or comment from you!

© 2011-2021 Zookal Pty Ltd

View question and answer

From our collection of questions and answers
Business · Accounting
Question details

I've attached what I have thus far and then the notes that I received back below the questions.

I need to prove A1 - A10. Some require specific numeric entries.

I attempted this but need to more information.

Given: A set R with two operations and true is a ring if the following eight properties are shown to be .Closure property of addition: for all s and t in R, s +t is also in R .Closure property of multiplication: for all s and t in R, s *t is also in R Additive identity property: there exists an element 0 in R such that s0 s for all s in Additive inverse property: for every s in R, there exists t in R, such that s+t0 Associative property of addition: for every q, s, and t in R, q (s + t)-(q + s) + t Associative property of multiplication: for every q, s, and t in R .Commutative property of addition: for all s and t in R, s +t-t+ s Left distributive property of multiplication over addition: for every q, s, and t in R q(s t)qs+q*t Right distributive property of multiplication over addition: for every q, s, and t in Rset of even integers. M22 la.bc,and d 2z) Recall that matrix addition and multiplication are defined as follows for 2 x 2 matrices: c+y d+ z The even integers 2Z form a ring with the usual operations of integer addition and multiplication. Given this fact, you are asked to prove that M(22) also has properties of a ring in part A. Each step of each proof must be justified using the definitions of the operations or an appropriate property from the even integers. The proofs of multiplicative closure (property 2) and associativity of multiplication (property 6) are provided for reference in the attached document Proof of Properties 2 and 6.A. Using the given definitions of the properties of a ring and given that 2Z is a ring, prove that M(22) with the operations of matrix addition and multiplication is also a ring by doing the following: 1. Prove the closure property of addition for M(2Z) 2. Prove the additive identity property for M(22) 3. Prove the additive inverse property for M(22) 4. Demonstrate the additive inverse property for M(22) by choosing elements A and -A in M(22 and verifying that A + -A are not all zero 0. The matrix A must have specific numeric entries that 5. Prove the associative property of addition for M(22) 6. Demonstrate the associative property of addition for M(22) by choosing elements A, B, and C in M(2z) and verifying that A + (B + C) (A + B) have specific numeric entries that are not all zero C. The matrices A, B, and C must 7. Prove the commutative property of addition for M(22) 8. Demonstrate the commutative property of addition for M(22) by choosing elements A and B in M(22) and verifying that A+ B-B + A. The matrices A and B must have specific numeric entries that are not all zero 9. Prove the left distributive property of multiplication over addition or the right distributive roperty of multiplication over addition for M(2Note: You are not required to prove both of the distributive properties. 10. Demonstrate one of the two distributive properties (right or left) by choosing elements A, B, and C in M(2 and verifying the property. The matrices A, B, and C must have specific numeric entries that are not all zero.

This is what I have, please be explicit in your response.

A2. The Additive Identity Property Since 0is an even integer,-0EMzz) E M (22) Then, A+0-2 bl. ro Therefore, the identity property holds in M (2Z)

1/7/19 Steps are shown leading from A+0 to A. The sequence of steps is incomplete, and no justifications were evident for the steps. To begin the demonstration, it is claimed that 0 is an even integer and the zero matrix is a member of M(2Z). Sufficient justification for these claims was not evident. The notation used for the integer 0 and the zero matrix is also identical. A clear distinction is needed.

A3. The Additive Inverse Property Let B be the additive inverse of A. Therefore, A + B = 0 a + w = 0, b + x = 0, c + y 0, d + z = 0 Therefore, B- a-b So every element of M (2Z) has an additive inverse in M (2Z)

1/7/19 Steps are shown which result in a matrix B such that A+B = 0. The steps shown and the existence of such a matrix in M(2Z) have not been justified. In the work to determine the entries of B, it is unclear what steps/operations were used to conclude that "w = -a, x = -b, y = -c, z = -d". Only the defined ring operations of addition and multiplication on 2Z and M(2Z) may be used in the development.

A4. The Associative Property for+ Let A, B, C E M (2Z) and A-a dl.B-ly ! Then, A + (B + C) = A+lytr z+1-le+y+r d+2+s! Now, (A B) +C So, A(BC)(AB) C Therefore, the associative property holds for M (2Z) for addition.

1/7/19 A specific illustration of the additive inverse property was not evident.

A5. The Commutative Property for+ Let A, BE M (2Z) and A-la dl. Now, A +B-a B-ly :Iwhere a, b, c, d, w, x, y, z are є (22) w .is. h w.1.殴 五4. Ь Therefore,Ё..+d-Ё. m since addition is commutative in e+ y d + zl the set of integers. Hence, A+B B+ A Thus, the commutative property holds for addition in M (2Z)

1/7/19 Some steps are included that address A + (B + C) and (A + B) + C. The steps are incomplete and no justification is evident. The entries of the final matrices shown for each side, such as "𝑎 + 𝑤 + 𝑝" lack proper grouping symbols to indicate the order of operations. Once these grouping symbols are included, additional steps and justification will be necessary to reconcile the two sides.

A6. The Left Distributuity Property of x Over Claim: AB+C ABAC where A, B,CEM (22). where a, b, c,d,w, x, Y, 2, P, Q.,sare E (22) alw + p) +byr)ax+ b(z+s) + dy cx+ dz +br at + dr +ds С.low+by axthHE + br aq Then, A A-8+Ady cx+dz Therefore, addition is commutative in (22) Hence, AIB+C)-ABAC

1/7/19 A specific illustration of the additive associative property was not evident.

Answer
Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.

Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.Find step-by-step answers from expert tutors to questions asked by students like you. Start 14-day free trial.