EBK ALGEBRA FOUNDATIONS
EBK ALGEBRA FOUNDATIONS
15th Edition
ISBN: 9780321978929
Author: Martin-Gay
Publisher: PEARSON
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Chapter F, Problem 1ES

Evaluate. See Example 1.

  1. 3 5 1 7

Expert Solution & Answer
Check Mark
To determine

To evaluate:the givenexpression using matrices properties and mathematical methodologies.

Answer to Problem 1ES

The value of given expression in simplified form is =26 .

Explanation of Solution

Given information:

The given is below expression:

  |3 51 7|

Calculation:

The given is below expression:

  |3 51 7|

The given is in matrices form and required is its determinant, calculated as below:

  |3 51 7|=(3)(7)(5)(1)=21+5=26

Hence, the value of given expression in simplified form is =26 .

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************* ********************************* Q.1) Classify the following statements as a true or false statements: a. If M is a module, then every proper submodule of M is contained in a maximal submodule of M. b. The sum of a finite family of small submodules of a module M is small in M. c. Zz is directly indecomposable. d. An epimorphism a: M→ N is called solit iff Ker(a) is a direct summand in M. e. The Z-module has two composition series. Z 6Z f. Zz does not have a composition series. g. Any finitely generated module is a free module. h. If O→A MW→ 0 is short exact sequence then f is epimorphism. i. If f is a homomorphism then f-1 is also a homomorphism. Maximal C≤A if and only if is simple. Sup Q.4) Give an example and explain your claim in each case: Monomorphism not split. b) A finite free module. c) Semisimple module. d) A small submodule A of a module N and a homomorphism op: MN, but (A) is not small in M.
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