031 Review Part 3, Problem 1 - Revision history

Kayla Murray at 20:51, 15 October 2017

2017-10-15T20:51:47Z

Kayla Murray at 00:09, 14 October 2017

2017-10-14T00:09:20Z

Kayla Murray at 00:05, 14 October 2017

2017-10-14T00:05:02Z

Kayla Murray at 23:35, 13 October 2017

2017-10-13T23:35:07Z

Kayla Murray at 23:15, 13 October 2017

2017-10-13T23:15:55Z

Kayla Murray at 20:10, 13 October 2017

2017-10-13T20:10:33Z

Kayla Murray at 02:21, 10 October 2017

2017-10-10T02:21:48Z

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

2017-10-10T02:02:49Z

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:  
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|   '''(a)'''    
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|   '''(b)'''    
|}
[[031_Review_Part_3|'''Return to Sample Exam''']]

@@ Line 11: / Line 11: @@
 & 0 & 3
           \end{bmatrix}</math>&nbsp; diagonalizable? If so, explain why and diagonalize it. If not, explain why not.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 164: / Line 163: @@
 |-
+|
-::<math>\bigg\{\begin{bmatrix}
+::<math>\Bigg\{\begin{bmatrix}
             -1  \\
 \\
-          \end{bmatrix}\bigg\}.</math>
+          \end{bmatrix}\Bigg\}.</math>
 |}
@@ Line 204: / Line 203: @@
 |-
+|
-::<math>\bigg\{\begin{bmatrix}
+::<math>\Bigg\{\begin{bmatrix}
   \\
 \\
@@ Line 212: / Line 211: @@
 \\
-          \end{bmatrix}\bigg\}.</math>
+          \end{bmatrix}\Bigg\}.</math>
 |}
@@ Line 253: / Line 252: @@
           \end{bmatrix}.</math>
 |}
-[[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]
+[[031_Review_Part_3|'''<u>Return to Review Problems</u>''']]

@@ Line 174: / Line 174: @@
 !Step 3: &nbsp;
 |-
-|For the eigenvalue &nbsp;<math>\lambda=3,</math>&nbsp; we have
+|For the eigenvalue &nbsp;<math style="vertical-align: -4px">\lambda=3,</math>&nbsp; we have
 |-
+|
@@ Line 201: / Line 201: @@
 \end{array}</math>
 |-
-|We see that &nbsp;<math>x_2</math>&nbsp; and &nbsp;<math>x_3</math>&nbsp; are free variables. So, a basis for the eigenspace corresponding to &nbsp;<math>3</math>&nbsp; is
+|We see that &nbsp;<math style="vertical-align: -3px">x_2</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">x_3</math>&nbsp; are free variables. So, a basis for the eigenspace corresponding to &nbsp;<math style="vertical-align: -1px">3</math>&nbsp; is
 |-
+|
@@ Line 218: / Line 218: @@
 !Step 4: &nbsp;
 |-
-|Since &nbsp;<math>A</math>&nbsp; has &nbsp;<math>3</math>&nbsp; linearly independent eigenvectors,
+|Since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; has &nbsp;<math style="vertical-align: 0px">3</math>&nbsp; linearly independent eigenvectors, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is diagonalizable by the Diagonalization Theorem.
 |-
-|<math>A</math>&nbsp; is diagonalizable by the Diagonalization Theorem.
+|Using the Diagonalization Theorem, we can diagonalize &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; using the information from the steps above.
 |-
 |So, we have

@@ Line 131: / Line 131: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |}
@@ Line 142: / Line 245: @@
 |-
-|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: 0px">A</math>&nbsp; is diagonalizable and &nbsp;<math>D=\begin{bmatrix}
 |}
 [[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 90: / Line 90: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
 |-
+|
 |}

@@ Line 34: / Line 34: @@
 !Step 1: &nbsp;
 |-
+|To answer this question, we examine the eigenvalues and eigenvectors of &nbsp;<math style="vertical-align: 0px">A.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |}
@@ Line 61: / Line 104: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: 0px">A</math>&nbsp; is not diagonalizable.
 |-
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;
 |}
 [[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 15: / Line 15: @@
 {| 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">(a) Is the matrix &nbsp;<math style="vertical-align: -18px">A=
      \begin{bmatrix}
-& -4 & 9 & -7 \\
+& 1 \\
-           -1 & 2  & -4 & 1 \\
+& 3
-& -6 & 10 & 7
+          \end{bmatrix}</math>&nbsp; diagonalizable? If so, explain why and diagonalize it. If not, explain why not.
-          \end{bmatrix}</math>&nbsp; and assume that it is row equivalent to the matrix
+<span class="exam">(b) Is the matrix &nbsp;<math style="vertical-align: -31px">A=
      \begin{bmatrix}
-& 0 & -1 & 5 \\
+& 0 & -2 \\
-& -2  & 5 & -6 \\
+& 3  & 2 \\
-& 0 & 0 & 0
+& 0 & 3
-          \end{bmatrix}.</math>
+          \end{bmatrix}</math>&nbsp; diagonalizable? If so, explain why and diagonalize it. If not, explain why not.