/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q67E Consider a pattern聽聽P聽 in a聽... [FREE SOLUTION] | 91影视

91影视

Consider a patternP in ann matrix, and choose an entryaijin this pattern. Show that the number of inversions involvingaij is even if (i+j) is even and odd if (i+j) is odd. Hint: Suppose there arek entries in the pattern to the left and above aij . Express the number of inversions involvingaij in terms ofk.

Short Answer

Expert verified

Therefore, the number of inversions including aijis i-1-k+j-1-k=i+j-2-2k, which is even if i+j is even, and odd if i+j is odd.

Step by step solution

01

Matrix Definition 

A matrix is a set of numbers arranged in rows and columns so as to form a rectangulararray.

The numbers are called the elements, or entries, of the matrix.

If there are m rows and n columns, the matrix is said to be a 鈥m by n 鈥 matrix, written 鈥 mn.鈥

02

To find the number of inversions involving  aij in terms of k

Consider k to be the number of entries, in this pattern, that are to the left and above aij.

Clearly then, the number of the pattern's entries to the right and above aijis i-1-k.

In a similar manner, the number of such entries to the left and below aij is j-1-k.

Therefore, the number of inversions including aijisrole="math" localid="1660723822760" i-1-k+j-1-k=i+j-2-2k, which is even if i+j is even, and odd if i+j is odd.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.