Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 6E from Chapter 2.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 6E
Chapter:
Problem:
In Exercises, compute the product AB in two ways:(a) by the definition, where Ab1 and Ab2 are computed separately, and(b) by the row-column rule for computing AB.
Step-by-Step Solution
Given Information
we are given with following Matrices:\[A = \left[ {\begin{array}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}1&3\\2&{ - 1}\end{array}} \right]\]We have to determine the product of these matrices (a) by the definition, where $Ab_1$ and $Ab_2$ are computed separately, and (b) by the row–column rule for computing AB.
Step 1: Part (a)
\[\begin{array}{l}A{b_1} = \left[ {\begin{array}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{array}} \right]\left[ {\begin{array}{*{20}{l}}1\\2\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{4 \times 1 + ( - 2) \times 2}\\{( - 3) \times 1 + 0 \times 2}\\{3 \times 1 + 5 \times 2}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}0\\{ - 3}\\{13}\end{array}} \right]\end{array}\]Therefore \[A{b_1} = \left[ {\begin{array}{*{20}{c}}0\\{ - 3}\\{13}\end{array}} \right]\]
Step 2:
\[\begin{array}{l}A{b_2} = \left[ {\begin{array}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3\\{ - 1}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{4 \times 3 + ( - 2) \times ( - 1)}\\{( - 3) \times 3 + 0 \times ( - 1)}\\{3 \times 3 + 5 \times ( - 1)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{14}\\{ - 9}\\4\end{array}} \right]\end{array}\]Therefore \[A{b_2} = \left[ {\begin{array}{*{20}{c}}{14}\\{ - 9}\\4\end{array}} \right]\]Therefore, the product AB is \[AB = \left[ {A{b_1}A{b_2}} \right] = \left[ {\begin{array}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{array}} \right]\]
Step 3: (b)
\[\begin{array}{l}AB = \left[ {\begin{array}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&3\\2&{ - 1}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{4 \times 1 + ( - 2) \times 2}&{4 \times 3 + ( - 2) \times ( - 1)}\\{( - 3) \times 1 + 0 \times 2}&{( - 3) \times 3 + 0 \times ( - 1)}\\{3 \times 1 + 5 \times 2}&{3 \times 3 + 5 \times ( - 1)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{array}} \right]\end{array}\]Therefore, the product AB is \[AB = \left[ {\begin{array}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{array}} \right]\]
we are given with following Matrices:\[A = \left[ {\begin{array}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}1&3\\2&{ - 1}\end{array}} \right]\]We have to determine the product of these matrices (a) by the definition, where $Ab_1$ and $Ab_2$ are computed separately, and (b) by the row–column rule for computing AB.
Step 1: Part (a)
\[\begin{array}{l}A{b_1} = \left[ {\begin{array}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{array}} \right]\left[ {\begin{array}{*{20}{l}}1\\2\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{4 \times 1 + ( - 2) \times 2}\\{( - 3) \times 1 + 0 \times 2}\\{3 \times 1 + 5 \times 2}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}0\\{ - 3}\\{13}\end{array}} \right]\end{array}\]Therefore \[A{b_1} = \left[ {\begin{array}{*{20}{c}}0\\{ - 3}\\{13}\end{array}} \right]\]
Step 2:
\[\begin{array}{l}A{b_2} = \left[ {\begin{array}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3\\{ - 1}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{4 \times 3 + ( - 2) \times ( - 1)}\\{( - 3) \times 3 + 0 \times ( - 1)}\\{3 \times 3 + 5 \times ( - 1)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{14}\\{ - 9}\\4\end{array}} \right]\end{array}\]Therefore \[A{b_2} = \left[ {\begin{array}{*{20}{c}}{14}\\{ - 9}\\4\end{array}} \right]\]Therefore, the product AB is \[AB = \left[ {A{b_1}A{b_2}} \right] = \left[ {\begin{array}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{array}} \right]\]
Step 3: (b)
\[\begin{array}{l}AB = \left[ {\begin{array}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&3\\2&{ - 1}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{4 \times 1 + ( - 2) \times 2}&{4 \times 3 + ( - 2) \times ( - 1)}\\{( - 3) \times 1 + 0 \times 2}&{( - 3) \times 3 + 0 \times ( - 1)}\\{3 \times 1 + 5 \times 2}&{3 \times 3 + 5 \times ( - 1)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{array}} \right]\end{array}\]Therefore, the product AB is \[AB = \left[ {\begin{array}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{array}} \right]\]