031 Review Part 2, Problem 1 - Revision history

Kayla Murray at 20:12, 15 October 2017

2017-10-15T20:12:10Z

Kayla Murray at 00:56, 11 October 2017

2017-10-11T00:56:46Z

Kayla Murray at 00:23, 11 October 2017

2017-10-11T00:23:48Z

Kayla Murray at 17:34, 10 October 2017

2017-10-10T17:34:59Z

Kayla Murray at 17:22, 10 October 2017

2017-10-10T17:22:39Z

Kayla Murray at 17:14, 10 October 2017

2017-10-10T17:14:41Z

Kayla Murray at 01:56, 10 October 2017

2017-10-10T01:56:20Z

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

2017-10-10T01:55:01Z

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)'''    
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|   '''(b)'''    
|}
[[009C_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 16: / Line 16: @@
 <span class="exam">(b) Find bases for &nbsp;<math style="vertical-align: 0px">\text{Col }A</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">\text{Nul }A.</math>&nbsp; Find an example of a nonzero vector that belongs to &nbsp;<math style="vertical-align: -5px">\text{Col }A,</math>&nbsp; as well as an example of a nonzero vector that belongs to &nbsp;<math style="vertical-align: 0px">\text{Nul }A.</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 132: / Line 131: @@
           \end{bmatrix}.</math>
 |-
-|Hence, a basis for &nbsp;<math style="vertical-align: -1px">\text{Nul }A</math>&nbsp; is
+|Therefore, a basis for &nbsp;<math style="vertical-align: -1px">\text{Nul }A</math>&nbsp; is
 |-
+|
@@ Line 182: / Line 181: @@
           </math>
 |}
-[[031_Review_Part_2|'''<u>Return to Sample Exam</u>''']]
+[[031_Review_Part_2|'''<u>Return to Review Problems</u>''']]

@@ Line 95: / Line 95: @@
 !Step 2: &nbsp;
 |-
-|To find a basis for &nbsp;<math>\text{Nul }A</math> we translate the matrix equation &nbsp;<math>Bx=0</math>&nbsp; back into a system of equations
+|To find a basis for &nbsp;<math style="vertical-align: -1px">\text{Nul }A</math>&nbsp; we translate the matrix equation &nbsp;<math style="vertical-align: -1px">Bx=0</math>&nbsp; back into a system of equations
 |-
 |and solve for the pivot variables.
@@ Line 117: / Line 117: @@
 \end{array}</math>
 |-
-|Hence, the solutions to &nbsp;<math>Ax=0</math>&nbsp; are of the form
+|Hence, the solutions to &nbsp;<math style="vertical-align: -1px">Ax=0</math>&nbsp; are of the form
 |-
+|
@@ Line 154: / Line 154: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; <math>\text{rank }A=2</math>&nbsp; and <math style="vertical-align: -2px">\text{dim Nul }A=2.</math>
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -2px">\text{rank }A=2</math>&nbsp; and &nbsp;<math style="vertical-align: -2px">\text{dim Nul }A=2</math>
 |-
-|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; A basis for &nbsp;<math>\text{Col }A</math>&nbsp; is <math>\Bigg\{\begin{bmatrix}
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; A basis for &nbsp;<math style="vertical-align: -1px">\text{Col }A</math>&nbsp; is &nbsp;<math>\Bigg\{\begin{bmatrix}
   \\
             -1 \\
@@ Line 165: / Line 165: @@
 \\
             -6
-          \end{bmatrix}\Bigg\}.
+          \end{bmatrix}\Bigg\}
           </math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; and a basis for &nbsp;<math>\text{Nul }A</math>&nbsp; is <math>\Bigg\{\begin{bmatrix}
+|&nbsp; &nbsp; &nbsp; &nbsp; and a basis for &nbsp;<math style="vertical-align: -1px">\text{Nul }A</math>&nbsp; is &nbsp;<math>\Bigg\{\begin{bmatrix}
   \\
             \frac{5}{2} \\

@@ Line 94: / Line 94: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |}
@@ Line 102: / Line 154: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; <math>\text{rank }A=2</math>&nbsp; and <math style="vertical-align: -2px">\text{dim Nul }A=2.</math>
 |-
-|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;
+|&nbsp; &nbsp; &nbsp; &nbsp; and a basis for &nbsp;<math>\text{Nul }A</math>&nbsp; is <math>\Bigg\{\begin{bmatrix}
 |}
 [[031_Review_Part_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 63: / Line 63: @@
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;
 |}
-[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
+[[031_Review_Part_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 70: / Line 70: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
 |-
+|
 |}

@@ Line 20: / Line 20: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
+|
 |}