031 Review Part 2, Problem 6 - Revision history

Kayla Murray at 20:34, 15 October 2017

2017-10-15T20:34:46Z

Kayla Murray at 22:24, 12 October 2017

2017-10-12T22:24:35Z

Kayla Murray at 22:04, 12 October 2017

2017-10-12T22:04:25Z

Kayla Murray at 03:55, 12 October 2017

2017-10-12T03:55:35Z

Kayla Murray at 02:10, 10 October 2017

2017-10-10T02:10:44Z

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

2017-10-10T02:01:44Z

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)'''    
|}
[[031_Review_Part_2|'''Return to Sample Exam''']]

@@ Line 14: / Line 14: @@
 <span class="exam">(c) Let &nbsp;<math style="vertical-align: -5px">L=\text{Span }\{\vec{v}\}.</math>&nbsp; Compute the orthogonal projection of &nbsp;<math style="vertical-align: -3px">\vec{y}</math>&nbsp; onto &nbsp;<math style="vertical-align: 0px">L.</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 27: / Line 26: @@
 |-
+|
-::<math>\hat{y}=\text{proj}_L \vec{y}=\frac{\vec{y}\cdot \vec{u}}{\vec{u}\cdot \vec{u}}\vec{u}.</math>
+::<math>\hat{y}=\text{proj}_L \vec{y}=\frac{\vec{y}\cdot \vec{v}}{\vec{v}\cdot \vec{v}}\vec{v}.</math>
 |}
@@ Line 141: / Line 140: @@
 &&\\
 & = & \displaystyle{\begin{bmatrix}
-            \frac{1}{5} \\
+            \frac{1}{10} \\
-            -\frac{3}{5} \\
+            \frac{-3}{10} \\
           \end{bmatrix}.}
@@ Line 161: / Line 160: @@
 |-
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math>\begin{bmatrix}
-            \frac{1}{5} \\
+            \frac{1}{10} \\
-            -\frac{3}{5} \\
+            \frac{-3}{10} \\
           \end{bmatrix}</math>
 |}
-[[031_Review_Part_2|'''<u>Return to Sample Exam</u>''']]
+[[031_Review_Part_2|'''<u>Return to Review Problems</u>''']]

@@ Line 38: / Line 38: @@
 !Step 1: &nbsp;
 |-
-|First, we calculate &nbsp;<math>||\vec{v}||.</math>&nbsp;
+|First, we calculate &nbsp;<math style="vertical-align: -4px">||\vec{v}||.</math>&nbsp;
 |-
 |We get
@@ Line 55: / Line 55: @@
 !Step 2: &nbsp;
 |-
-|Now, to get a unit vector in the direction of &nbsp;<math>\vec{v},</math>&nbsp; we take the vector &nbsp;<math>\vec{v}</math>&nbsp; and divide by &nbsp;<math>||\vec{v}||.</math>
+|Now, to get a unit vector in the direction of &nbsp;<math style="vertical-align: -4px">\vec{v},</math>&nbsp; we take the vector &nbsp;<math style="vertical-align: 0px">\vec{v}</math>&nbsp; and divide by &nbsp;<math style="vertical-align: -4px">||\vec{v}||.</math>
 |-
 |Hence, we get the vector
@@ Line 78: / Line 78: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
 |-
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |}
@@ Line 92: / Line 112: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
 |-
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |}
@@ Line 106: / Line 152: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; <math>\begin{bmatrix}
 |-
-|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math>\begin{bmatrix}
 |}
 [[031_Review_Part_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 37: / Line 37: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
 |-
+|
 |}
@@ Line 44: / Line 55: @@
 !Step 2: &nbsp;
 |-
+|Now, to get a unit vector in the direction of &nbsp;<math>\vec{v},</math>&nbsp; we take the vector &nbsp;<math>\vec{v}</math>&nbsp; and divide by &nbsp;<math>||\vec{v}||.</math>
 |}

@@ Line 18: / Line 18: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
+|
 |}

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the matrix &nbsp;<math style="vertical-align: -31px">A=
+<span class="exam"> Let &nbsp;<math>\vec{v}=\begin{bmatrix}
-    \begin{bmatrix}
+            -1 \\
-& -4 & 9 & -7 \\
+\\
-            -1 & 2  & -4 & 1 \\
-& -6 & 10 & 7
+          \end{bmatrix}</math>&nbsp; and &nbsp;<math>\vec{y}=\begin{bmatrix}
-          \end{bmatrix}</math>&nbsp; and assume that it is row equivalent to the matrix
+\\
 \\
-::<math>B=
-    \begin{bmatrix}
+          \end{bmatrix}.</math>
-& 0 & -1 & 5 \\
-& -2  & 5 & -6 \\
+<span class="exam">(a) Find a unit vector in the direction of &nbsp;<math style="vertical-align: 0px">\vec{v}.</math>
-& 0 & 0 & 0
-          \end{bmatrix}.</math>
+<span class="exam">(b) Find the distance between &nbsp;<math style="vertical-align: 0px">\vec{v}</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">\vec{y}.</math>
-−
-<span class="exam">(a) List rank &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">\text{dim Nul }A.</math>
+<span class="exam">(c) Let &nbsp;<math style="vertical-align: -5px">L=\text{Span }\{\vec{v}\}.</math>&nbsp; Compute the orthogonal projection of &nbsp;<math style="vertical-align: -3px">\vec{y}</math>&nbsp; onto &nbsp;<math style="vertical-align: 0px">L.</math>
@@ Line 42: / Line 40: @@
 '''(b)'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"