/*! 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} Q16Q Suppose A聽and B聽are \(n \times... [FREE SOLUTION] | 91影视

91影视

Suppose Aand Bare \(n \times n\), Bis invertible, and ABis invertible. Show that Ais invertible. (Hint: Let C=AB, and solve this equation for A.)

Short Answer

Expert verified

It is proved that Ais invertible.

Step by step solution

01

Condition for an invertible matrix

Theorem 5states that Ais an invertible \(n \times n\) matrix, then for each b in \({\mathbb{R}^n}\), the equation \(Ax = b\) has a unique solution \(x = {A^{ - 1}}b\).

02

Show that A is invertible

Theorem 6states that Aand Bare \(n \times n\) invertible matrices; the inverse of ABis the product of the inverse of A and B in the reverse order. That is, \({\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}\).

Consider \(C = AB\).

Multiply each side of the equation \(C = AB\) by \({B^{ - 1}}\):

\(\begin{aligned}{c}C{B^{ - 1}} = AB{B^{ - 1}}\\ = AI\\ = A\end{aligned}\)

According to theorem 6, Ais the product of invertible matrices; thus, Ais invertible.

Hence, it is proved that Ais invertible.

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

In exercise 11 and 12, the matrices are all \(n \times n\). Each part of the exercise is an implication of the form 鈥淚f 鈥渟tatement 1鈥 then 鈥渟tatement 2鈥.鈥滿ark the implication as True if the truth of 鈥渟tatement 2鈥漚lways follows whenever 鈥渟tatement 1鈥 happens to be true. An implication is False if there is an instance in which 鈥渟tatement 2鈥 is false but 鈥渟tatement 1鈥 is true. Justify each answer.

a. If the equation \[A{\bf{x}} = {\bf{0}}\] has only the trivial solution, then \(A\) is row equivalent to the \(n \times n\) identity matrix.

b. If the columns of \(A\) span \({\mathbb{R}^n}\), then the columns are linearly independent.

c. If \(A\) is an \(n \times n\) matrix, then the equation \(A{\bf{x}} = {\bf{b}}\) has at least one solution for each \({\bf{b}}\) in \({\mathbb{R}^n}\).

d. If the equation \[A{\bf{x}} = {\bf{0}}\] has a non trivial solution, then \[A\] has fewer than \(n\) pivot positions.

e. If \({A^T}\) is not invertible, then \(A\) is not invertible.

Exercises 23-26 display a matrix A and echelon form of A. Find a basis for Col A and a basis for Nul A.

\[A = \left[ {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{5}}&{\bf{9}}&{ - {\bf{2}}}\\{\bf{6}}&{\bf{5}}&{\bf{1}}&{{\bf{12}}}\\{\bf{3}}&{\bf{4}}&{\bf{8}}&{ - {\bf{3}}}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{6}}&{ - {\bf{5}}}\\{\bf{0}}&{\bf{1}}&{\bf{5}}&{ - {\bf{6}}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\end{array}} \right]\]

Unless otherwise specified, assume that all matrices in these exercises are \(n \times n\). Determine which of the matrices in Exercises 1-10 are invertible. Use a few calculations as possible. Justify your answer.

10. [M] \[\left[ {\begin{array}{*{20}{c}}5&3&1&7&9\\6&4&2&8&{ - 8}\\7&5&3&{10}&9\\9&6&4&{ - 9}&{ - 5}\\8&5&2&{11}&4\end{array}} \right]\]

Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

Exercises 1-4 display sets in \({\mathbb{R}^2}\). Assume the sets include the bounding lines. In each case, give a specific reason why the set H is not a subspace of \({\mathbb{R}^2}\). (For instance, find two vectors in H whose sum is not in H, or find a vector in H with a scalar multiple that is not in H. Draw a picture.)

2.

See all solutions

Recommended explanations on Math Textbooks

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.