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 16E from Chapter 2.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 16E
Chapter:
Problem:
Exercises concern arbitrary matrices A, B, and C for which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.a. If A and B are 3 × 3 and
Step-by-Step Solution
Given Information
We have to marks the given statements as True or False and justify for the same.
Step 1: (a)
Statement: If A and B are $3 \times 3$ and $B = \left[ {{{\bf{b}}_{\bf{1}}}\,\,{{\bf{b}}_{\bf{2}}}\,\,{{\bf{b}}_{\bf{3}}}} \right]$; then $AB = \left[ {A{{\bf{b}}_{\bf{1}}} + A{{\bf{b}}_{\bf{2}}}\, + A{{\bf{b}}_{\bf{3}}}} \right]$.
The Statement is False! Here the $b_i$s are columns of size $3 \times 1$ and order of matrix A is $3 \times 3$, hence the product $Ab_1$ has order of $3 \times 1$. So, the order of $\left[ {A{{\bf{b}}_{\bf{1}}} + A{{\bf{b}}_{\bf{2}}}\, + A{{\bf{b}}_{\bf{3}}}} \right]$ is also $3 \times 1$. However, the product of two matrices of size $3 \times 3$ should also have order of $3 \times 3$. This is not the case here. So the Statement is
FALSE
Step 2: (b)
Statement: The second row of AB is the second row of A multipliedon the right by B.
The Statement is True! The statement is True from the Row-Column Law.
Therefore TRUE
Step 3: (c)
Statement: $(AB)C = (AC)B$
The Statement is False! Matrices do not hold Commutative property.
Therefore FALSE
Step 4: (d)
Statement: ${(AB)^T} = {A^T}{B^T}$.
The Statement is False! By the properties of the transposition, \[{(AB)^T} = {B^T}{A^T}\]
Therefore FALSE
Step 5: (e)
Statement: The transpose of a sum of matrices equals the sum of theirtransposes..
The Statement is TRUE! By the properties of the transposition,
Therefore TRUE
We have to marks the given statements as True or False and justify for the same.
Step 1: (a)
Statement: If A and B are $3 \times 3$ and $B = \left[ {{{\bf{b}}_{\bf{1}}}\,\,{{\bf{b}}_{\bf{2}}}\,\,{{\bf{b}}_{\bf{3}}}} \right]$; then $AB = \left[ {A{{\bf{b}}_{\bf{1}}} + A{{\bf{b}}_{\bf{2}}}\, + A{{\bf{b}}_{\bf{3}}}} \right]$.
The Statement is False! Here the $b_i$s are columns of size $3 \times 1$ and order of matrix A is $3 \times 3$, hence the product $Ab_1$ has order of $3 \times 1$. So, the order of $\left[ {A{{\bf{b}}_{\bf{1}}} + A{{\bf{b}}_{\bf{2}}}\, + A{{\bf{b}}_{\bf{3}}}} \right]$ is also $3 \times 1$. However, the product of two matrices of size $3 \times 3$ should also have order of $3 \times 3$. This is not the case here. So the Statement is
FALSE
Step 2: (b)
Statement: The second row of AB is the second row of A multipliedon the right by B.
The Statement is True! The statement is True from the Row-Column Law.
Therefore TRUE
Step 3: (c)
Statement: $(AB)C = (AC)B$
The Statement is False! Matrices do not hold Commutative property.
Therefore FALSE
Step 4: (d)
Statement: ${(AB)^T} = {A^T}{B^T}$.
The Statement is False! By the properties of the transposition, \[{(AB)^T} = {B^T}{A^T}\]
Therefore FALSE
Step 5: (e)
Statement: The transpose of a sum of matrices equals the sum of theirtransposes..
The Statement is TRUE! By the properties of the transposition,
Therefore TRUE