/*! 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} Q6Q In exercise 5 and 6, compute the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q6Q (page 93)

In exercise 5 and 6, compute the product \(AB\) in two ways: (a) by the definition, where \(A{b_{\bf{1}}}\) and \(A{b_{\bf{2}}}\) are computed separately, and (b) by the row-column rule for computing \(AB\).

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{2}}}\\{ - {\bf{3}}}&{\bf{0}}\\{\bf{3}}&{\bf{5}}\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{3}}\\{\bf{2}}&{ - {\bf{1}}}\end{aligned}} \right)\)

Short Answer

Expert verified

\(\left( {\begin{aligned}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{aligned}} \right)\)

Step by step solution

01

Find the value of \(A{b_{\bf{1}}}\)

Multiply matrix \(A\) with the first column of matrix \(B\).

\(\begin{aligned}{c}A{b_1} = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1\\2\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4 \times 1 + \left( { - 2} \right) \times 2}\\{\left( { - 3} \right) \times 1 + 0}\\{3 \times 1 + 5 \times 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0\\{ - 3}\\{13}\end{aligned}} \right)\end{aligned}\)

02

Find the value of \(A{b_{\bf{2}}}\)

Multiply matrix \(A\) with the second column of matrix \(B\).

\(\begin{aligned}{c}A{b_2} = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4 \times 3 + \left( { - 2} \right) \times \left( { - 1} \right)}\\{\left( { - 3} \right) \times 3 + 0}\\{3 \times 3 + 5 \times \left( { - 1} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{14}\\{ - 9}\\4\end{aligned}} \right)\end{aligned}\)

03

Write the product \(AB\)

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

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}{A{b_1}}&{A{b_2}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{aligned}} \right)\end{aligned}\)

04

Find the product \(AB\) using row-column rule

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&3\\2&{ - 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4 \times 1 + \left( { - 2} \right) \times 2}&{4 \times 3 + \left( { - 2} \right) \times \left( { - 1} \right)}\\{\left( { - 3} \right) \times 1 + 0}&{\left( { - 3} \right) \times 3 + 0}\\{3 \times 1 + 5 \times 2}&{3 \times 3 + 5 \times \left( { - 1} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{aligned}} \right)\end{aligned}\)

So, \(AB = \left( {\begin{aligned}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{aligned}} \right)\).

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

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

Exercises 9鈥�12 display a matrix Aand an echelon form of A. Find bases for Col Aand Nul A, and then state the dimensions of these subspaces.

\(A = \left[ {\begin{array}{*{20}{c}}1&2&{ - 4}&3&3\\5&{10}&{ - 9}&{ - 7}&8\\4&8&{ - 9}&{ - 2}&7\\{ - 2}&{ - 4}&5&0&{ - 6}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&2&{ - 4}&3&3\\0&0&1&{ - 2}&0\\0&0&0&0&{ - 5}\\0&0&0&0&0\end{array}} \right]\)a

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

The solution to the steady-state heat flow problem for
the plate in the figure is approximated by the solution to the
equation\(A{\bf{x}} = {\bf{b}}\);where\(b = \left( {5,15,0,10,0,10,20,30} \right)\)and

\(A = \left[ {\begin{array}{*{20}{c}}4&{ - 1}&{ - 1}&{}&{}&{}&{}&{}\\{ - 1}&4&0&{ - 1}&{}&{}&{}&{}\\{ - 1}&0&4&{ - 1}&{ - 1}&{}&{}&{}\\{}&{ - 1}&{ - 1}&4&0&{ - 1}&{}&{}\\{}&{}&{ - 1}&0&4&{ - 1}&{ - 1}&{}\\{}&{}&{}&{ - 1}&{ - 1}&4&0&{ - 1}\\{}&{}&{}&{}&{ - 1}&0&4&{ - 1}\\{}&{}&{}&{}&{}&{ - 1}&{ - 1}&4\end{array}} \right]\)

(Refer to Exercise 33 of Section 1.1.) The missing entries in Aare zeros. The nonzero entries of A lie within a band along the main diagonal. Such band matricesoccur in a variety of applications and often are extremely large (with thousands of rows and columns but relatively narrow bands).

  1. Use the method of Example 2 to construct an LU factorization of A, and note that both factors are band matrices (with two nonzero diagonals below or above the main diagonal). Compute \(LU - A\) to check your work.
  1. Use the LU factorization to solve\(A{\bf{x}} = {\bf{b}}\).
  1. Obtain \({A^{ - {\bf{1}}}}\) and note that\({A^{ - {\bf{1}}}}\) is a dense matrix with no
    band structure. When Ais large, LandUcan be stored in
    much less space than\({A^{ - {\bf{1}}}}\). This fact is another reason for
    preferring the LU factorization of Ato \({A^{ - {\bf{1}}}}\) itself.

Suppose A is invertible. Explain why \({A^T}A\) is also invertible. Then show that \({A^{ - {\bf{1}}}} = {\left( {{A^T}A} \right)^{ - {\bf{1}}}}{A^T}\).
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Statistics

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.