Chapter 13: Q 71. (page 1066)

Prove Theorem $13.2 (a)$ . That is, prove that $\sum_{j = 1}^{m} \sum_{k = 1}^{n} a_{j k} = \sum_{k = 1}^{n} \sum_{j = 1}^{m} a_{j k}$

Short Answer

Expert verified

This is proved using expansion of left hand side of the equation.

Step by step solution

Given Information

We need to prove that

$\sum_{j = 1}^{m} \sum_{k = 1}^{n} a_{j k} = \sum_{k = 1}^{n} \sum_{j = 1}^{m} a_{j k}$

Simplification

Expansion gives $\sum_{j = 1}^{m} \sum_{k = 1}^{n} a_{j k} = \sum_{j = 1}^{m} (\sum_{k = 1}^{n} a_{j k})$

$= \sum_{j = 1}^{m} (a_{j 1} + a_{j 2} + a_{j 3} + \dots + a_{j n})$

$= (a_{11} + a_{12} + a_{13} + \dots + a_{1 n}) + (a_{21} + a_{22} + a_{23} + \dots + a_{2 n}) +$

$+ (a_{31} + a_{32} + a_{33} + \dots + a_{3 n}) + \dots + (a_{m 1} + a_{m 2} + a_{m 3} + \dots + a_{m n})$

Grouping gives

$\sum_{j = 1}^{m} \sum_{k = 1}^{n} a_{j k} = (a_{11} + a_{12} + a_{13} + \dots + a_{1 n}) + (a_{21} + a_{22} + a_{23} + \dots + a_{2 n}) +$

$+ (a_{31} + a_{32} + a_{33} + \dots + a_{3 n}) + \dots + (a_{m 1} + a_{m 2} + a_{m 3} + \dots + a_{m n})$

$= (a_{11} + a_{21} + a_{31} + \dots + a_{m 1}) + (a_{12} + a_{22} + a_{32} + \dots + a_{m 2}) +$

$+ (a_{13} + a_{23} + a_{33} + \dots + a_{m 3}) + \dots + (a_{1 n} + a_{2 n} + a_{3 n} + \dots + a_{m n})$

$\sum_{j = 1}^{m} \sum_{k = 1}^{n} a_{j k} = (a_{11} + a_{21} + a_{31} + \dots + a_{m 1}) + (a_{12} + a_{22} + a_{32} + \dots + a_{m 2}) +$

$+ (a_{13} + a_{23} + a_{33} + \dots + a_{m 3}) + \dots + (a_{1 n} + a_{2 n} + a_{3 n} + \dots + a_{m n})$

$= \sum_{k = 1}^{n} (a_{1 k} + a_{2 k} + a_{3 k} + \dots + a_{m k})$

$= \sum_{k = 1}^{n} \sum_{j = 1}^{m} a_{j k}$

Hence $\sum_{j = 1}^{m} \sum_{k = 1}^{n} a_{j k} = \sum_{k = 1}^{n} \sum_{j = 1}^{m} a_{j k}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Prove Theorem $13.2 (a)$ . That is, prove that $\sum_{j = 1}^{m} \sum_{k = 1}^{n} a_{j k} = \sum_{k = 1}^{n} \sum_{j = 1}^{m} a_{j k}$

Short Answer

Step by step solution

Given Information

Simplification

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Calculus

Theoretical and Mathematical Physics

Applied Mathematics

Statistics

Logic and Functions

Study anywhere. Anytime. Across all devices.

Prove Theorem 13.2(a). That is, prove that∑j=1m∑k=1najk=∑k=1n∑j=1majk

Short Answer

Step by step solution

Given Information

Simplification

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Calculus

Theoretical and Mathematical Physics

Applied Mathematics

Statistics

Logic and Functions

Study anywhere. Anytime. Across all devices.

Prove Theorem $13.2 (a)$ . That is, prove that $\sum_{j = 1}^{m} \sum_{k = 1}^{n} a_{j k} = \sum_{k = 1}^{n} \sum_{j = 1}^{m} a_{j k}$