031 Review Part 3, Problem 9 - Revision history

Kayla Murray at 21:07, 15 October 2017

2017-10-15T21:07:58Z

Kayla Murray at 17:11, 13 October 2017

2017-10-13T17:11:03Z

Kayla Murray at 02:31, 10 October 2017

2017-10-10T02:31:01Z

Kayla Murray: Created page with "Consider the matrix $A= \begin{bmatrix} 1 & -4 & 9 & -7 \\ -1 & 2 & -4 & 1 \\..."$

2017-10-10T02:04:27Z

Created page with "Consider the matrix <math style="vertical-align: -31px">A= \begin{bmatrix} 1 & -4 & 9 & -7 \\ -1 & 2 & -4 & 1 \\..."

New page

Consider the matrix  <math style="vertical-align: -31px">A=
\begin{bmatrix}
1 & -4 & 9 & -7 \\
-1 & 2 & -4 & 1 \\
5 & -6 & 10 & 7
\end{bmatrix}</math>  and assume that it is row equivalent to the matrix

::<math>B=
\begin{bmatrix}
1 & 0 & -1 & 5 \\
0 & -2 & 5 & -6 \\
0 & 0 & 0 & 0
\end{bmatrix}.</math>

(a) List rank  <math style="vertical-align: 0px">A</math>  and  <math style="vertical-align: 0px">\text{dim Nul }A.</math>

(b) Find bases for  <math style="vertical-align: 0px">\text{Col }A</math>  and  <math style="vertical-align: 0px">\text{Nul }A.</math>  Find an example of a nonzero vector that belongs to  <math style="vertical-align: -5px">\text{Col }A,</math>  as well as an example of a nonzero vector that belongs to  <math style="vertical-align: 0px">\text{Nul }A.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
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|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    
|-
|   '''(b)'''    
|}
[[031_Review_Part_3|'''Return to Sample Exam''']]

@@ Line 1: / Line 1: @@
 <span class="exam">Assume &nbsp;<math style="vertical-align: 0px">A^2=I.</math>&nbsp; Find &nbsp;<math style="vertical-align: -1px">\text{Nul }A.</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 42: / Line 41: @@
           \end{bmatrix}\Bigg\}</math>
 |}
-[[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]
+[[031_Review_Part_3|'''<u>Return to Review Problems</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the matrix &nbsp;<math style="vertical-align: -31px">A=
+<span class="exam">Assume &nbsp;<math style="vertical-align: 0px">A^2=I.</math>&nbsp; Find &nbsp;<math style="vertical-align: -1px">\text{Nul }A.</math>
@@ Line 26: / Line 10: @@
 '''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 59: / Line 27: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; &nbsp; &nbsp;
 |}
 [[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 5: / Line 5: @@
 !Foundations: &nbsp;
 |-
+|Recall that the subspace &nbsp;<math style="vertical-align: -1px">\text{Nul }A</math>&nbsp; is the set of all solutions to &nbsp;<math style="vertical-align: 0px">A\vec{x}=\vec{0}.</math>
 |}
@@ Line 14: / Line 14: @@
 !Step 1: &nbsp;
 |-
+|Since &nbsp;<math style="vertical-align: -4px">A\cdot A=I,</math>&nbsp; we know that &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |}
@@ Line 27: / Line 37: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; &nbsp; &nbsp;
+|&nbsp;&nbsp; &nbsp; &nbsp; <math>\text{Nul }A=\Bigg\{\begin{bmatrix}
 |}
 [[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]

031 Review Part 3, Problem 9 - Revision history

Kayla Murray at 21:07, 15 October 2017

Kayla Murray at 17:11, 13 October 2017

Kayla Murray at 02:31, 10 October 2017

Kayla Murray: Created page with "Consider the matrix A= \begin{bmatrix} 1 & -4 & 9 & -7 \\ -1 & 2 & -4 & 1 \\..."

Kayla Murray: Created page with "Consider the matrix $A= \begin{bmatrix} 1 & -4 & 9 & -7 \\ -1 & 2 & -4 & 1 \\..."$