/*! 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} Q4SE Suppose \({A^n} = {\bf{0}}\) for... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q4SE (page 93)

Suppose \({A^n} = {\bf{0}}\) for some \({\bf{n}} > {\bf{1}}\). Find an inverse for \(I - A\).

Short Answer

Expert verified

The inverse of \(I - A\) is \(I + A + {A^2} + ... + {A^{n - 1}}\).

Step by step solution

01

Use the matrix computation in Exercise 3

That is \(\left( {I - A} \right)\left( {I + A + {A^2}} \right) = I - {A^3}\). In general,

\(\left( {I - A} \right)\left( {I + A + {A^2} + ... + {A^{n - 1}}} \right) = I - {A^n}\)

02

Suppose \({A^n} = {\bf{0}}\)

Substitute \({A^n} = 0\) in the above equation to obtain:

\(\left( {I - A} \right)\left( {I + A + {A^2} + ... + {A^{n - 1}}} \right) = I\)

This implies \(I - A\) is invertibleand \({\left( {I - A} \right)^{ - 1}} = I + A + {A^2} + ... + {A^{n - 1}}\)

03

Conclusion

Hence, theinverse of \(I - A\) is \(I + A + {A^2} + ... + {A^{n - 1}}\).

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

Most popular questions from this chapter

Prove the Theorem 3(d) i.e., \({\left( {AB} \right)^T} = {B^T}{A^T}\).

Suppose the third column of Bis the sum of the first two columns. What can you say about the third column of AB? Why?

Use matrix algebra to show that if A is invertible and D satisfies \(AD = I\) then \(D = {A^{ - {\bf{1}}}}\).

Solve the equation \(AB = BC\) for A, assuming that A, B, and C are square and Bis invertible.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{5}}\\{ - {\bf{3}}}&{\bf{1}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{5}}}\\{\bf{3}}&k\end{aligned}} \right)\). What value(s) of \(k\), if any will make \(AB = BA\)?
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.