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 24E from Chapter 1.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 24E
Chapter:
Problem:
a. Every matrix equation Ax = b corresponds to a vector equation with the same solution set.b. Any linear combination of vectors can always be written in the form Ax for a suitable matrix A
Step-by-Step Solution
Given Information
We are given with some statements that we have to prove whether they are True or False
Step-1: (a)
Statement: Every matrix equation $Ax = b$ corresponds to a vector equation with the same solution set.
If A is a $m \times n$ matrix with columns $a_1, a_2, ...a_n$, and if b is in $R^m$, the the equation $Ax=b$ has the same solution set as the vector equation $x_1a_1 + x_2a_2...+x_na_n = b$
The Statement is True
Step-2: (b)
Statement: Any linear combination of vectors can always be written in the form $Ax$ for a suitable matrix $A$ and vector $x$.The statement is True, for example let us choose a linear combination ${\rm{2}}{{\rm{V}}_1} + {V_2} + 3{V_3}$ the linear combination can be written as: \[{\rm{2}}{{\rm{V}}_1} + {V_2} + 3{V_3} = \left[ {\begin{array}{*{20}{c}}{{V_1}}&{{V_2}}&{{V_3}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2\\1\\3\end{array}} \right]\]
The Statement is True
Step-3: (c)
Statement: The solution set of a linear system whose augmented matrix is $\left[ {{{\bf{a}}_{\bf{1}}}\,\,{{\bf{a}}_{\bf{2}}}\,\,{{\bf{a}}_{\bf{3}}}\,\,{\bf{b}}} \right]$ is the same as the solutionset of $Ax=b$ if $A = \left[ {{{\bf{a}}_{\bf{1}}}\,\,{{\bf{a}}_{\bf{2}}}\,\,{{\bf{a}}_{\bf{3}}}} \right]$
The statement is True as proved in statement (1)
The Statement is True
Step-4: (d)
Statement: If the equation $Ax = b$ is inconsistent, then b is not in the set spanned by the columns of A.
If b is not in the set spanned by the columns of A, it means that b is not a linear combination of the columns of A
The Statement is True
Step-5: (e)
Statement: If the augmented matrix [A b] has a pivot position in every row, then the equation $Ax = b$ is inconsistent..
If the augmented matrix [A b] has a pivot position in every row, it is not necessary that the equation $Ax=b$ is consistent.
The Statement is False
Step-6: (f)
Statement: If A is an $m \times n$ matrix whose columns do not span $R^m$, then the equation $Ax = b$ is inconsistent for some b in $R^m$.
If the columns of A do not span $R^m$, then the equation $Ax=0$ has no solution. So
The Statement is True
We are given with some statements that we have to prove whether they are True or False
Step-1: (a)
Statement: Every matrix equation $Ax = b$ corresponds to a vector equation with the same solution set.
If A is a $m \times n$ matrix with columns $a_1, a_2, ...a_n$, and if b is in $R^m$, the the equation $Ax=b$ has the same solution set as the vector equation $x_1a_1 + x_2a_2...+x_na_n = b$
The Statement is True
Step-2: (b)
Statement: Any linear combination of vectors can always be written in the form $Ax$ for a suitable matrix $A$ and vector $x$.The statement is True, for example let us choose a linear combination ${\rm{2}}{{\rm{V}}_1} + {V_2} + 3{V_3}$ the linear combination can be written as: \[{\rm{2}}{{\rm{V}}_1} + {V_2} + 3{V_3} = \left[ {\begin{array}{*{20}{c}}{{V_1}}&{{V_2}}&{{V_3}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2\\1\\3\end{array}} \right]\]
The Statement is True
Step-3: (c)
Statement: The solution set of a linear system whose augmented matrix is $\left[ {{{\bf{a}}_{\bf{1}}}\,\,{{\bf{a}}_{\bf{2}}}\,\,{{\bf{a}}_{\bf{3}}}\,\,{\bf{b}}} \right]$ is the same as the solutionset of $Ax=b$ if $A = \left[ {{{\bf{a}}_{\bf{1}}}\,\,{{\bf{a}}_{\bf{2}}}\,\,{{\bf{a}}_{\bf{3}}}} \right]$
The statement is True as proved in statement (1)
The Statement is True
Step-4: (d)
Statement: If the equation $Ax = b$ is inconsistent, then b is not in the set spanned by the columns of A.
If b is not in the set spanned by the columns of A, it means that b is not a linear combination of the columns of A
The Statement is True
Step-5: (e)
Statement: If the augmented matrix [A b] has a pivot position in every row, then the equation $Ax = b$ is inconsistent..
If the augmented matrix [A b] has a pivot position in every row, it is not necessary that the equation $Ax=b$ is consistent.
The Statement is False
Step-6: (f)
Statement: If A is an $m \times n$ matrix whose columns do not span $R^m$, then the equation $Ax = b$ is inconsistent for some b in $R^m$.
If the columns of A do not span $R^m$, then the equation $Ax=0$ has no solution. So
The Statement is True