/*! 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} Q10Q Let \(A = \left( {\begin{aligned... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q10Q (page 93)

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{4}}}&{\bf{6}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{4}}\\{\bf{5}}&{\bf{5}}\end{aligned}} \right)\) and \(C = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}&{ - {\bf{2}}}\\{\bf{3}}&{\bf{1}}\end{aligned}} \right)\). Verfiy that \(AB = AC\) and yet \(B \ne C\).

Short Answer

Expert verified

\(AB = AC\) and \(B \ne C\).

Step by step solution

01

Find the product \(AB\)

The product \(AB\) can be calculated as follows:

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}2&{ - 3}\\{ - 4}&6\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}8&4\\5&5\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{2 \times 8 + \left( { - 3} \right) \times 5}&{2 \times 4 + \left( { - 3} \right) \times 5}\\{\left( { - 4} \right) \times 8 + 6 \times 5}&{\left( { - 4} \right) \times 4 + 6 \times 5}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&{ - 7}\\{ - 2}&{14}\end{aligned}} \right)\end{aligned}\)

02

Find the product \(BA\)

The product \(BC\) can be calculated as follows:

\(\begin{aligned}{c}AC = \left( {\begin{aligned}{*{20}{c}}2&{ - 3}\\{ - 4}&6\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}5&{ - 2}\\3&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{2 \times 5 + 3 \times \left( { - 3} \right)}&{2 \times \left( { - 2} \right) + \left( { - 3} \right) \times 1}\\{ - 4 \times 5 + 6 \times 3}&{\left( { - 4} \right) \times \left( { - 2} \right) + 6 \times 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&{ - 7}\\{ - 2}&{14}\end{aligned}} \right)\end{aligned}\)

03

Compare the products of \(AB\) and \(AC\)

On comparing \(AB\) and \(AC\),

\(AB = AC = \left( {\begin{aligned}{*{20}{c}}1&{ - 7}\\{ - 2}&{14}\end{aligned}} \right)\).

So, \(AB = AC\) when \(B \ne C\).

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

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

36. Write the command(s) that will create a \(6 \times 4\) matrix with random entries. In what range of numbers do the entries lie? Tell how to create a \(3 \times 3\) matrix with random integer entries between \( - {\bf{9}}\) and 9. (Hint:If xis a random number such that 0 < x < 1, then \( - 9.5 < 19\left( {x - .5} \right) < 9.5\).

Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{1}}}\\{\bf{5}}&{ - {\bf{2}}}\end{aligned}} \right)\). Compute \({\bf{3}}{I_{\bf{2}}} - A\) and \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\).

If A, B, and X are \(n \times n\) invertible matrices, does the equation \({C^{ - 1}}\left( {A + X} \right){B^{ - 1}} = {I_n}\) have a solution, X? If so, find it.

Use the inverse found in Exercise 1 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{6}}{{\bf{x}}_{\bf{2}}} = {\bf{2}}\\{\bf{5}}{{\bf{x}}_{\bf{1}}} + {\bf{4}}{{\bf{x}}_{\bf{2}}} = - {\bf{1}}\end{aligned}\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Mechanics 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.