031 Review Part 3, Problem 6 - Revision history

Kayla Murray at 21:02, 15 October 2017

2017-10-15T21:02:06Z

Kayla Murray at 15:34, 11 October 2017

2017-10-11T15:34:56Z

Kayla Murray at 15:27, 11 October 2017

2017-10-11T15:27:29Z

Kayla Murray at 04:05, 11 October 2017

2017-10-11T04:05:41Z

Kayla Murray at 02:25, 10 October 2017

2017-10-10T02:25:28Z

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

2017-10-10T02:03:49Z

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>

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!Foundations:  
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'''Solution:'''

'''(a)'''

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'''(b)'''

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!Final Answer:  
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|   '''(a)'''    
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|   '''(b)'''    
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[[031_Review_Part_3|'''Return to Sample Exam''']]

@@ Line 2: / Line 2: @@
 <span class="exam">(b) Show that if &nbsp;<math style="vertical-align: -3px">\vec{y}</math>&nbsp; is an eigenvector of the matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; corresponding to the eigenvalue 3 and &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible, then &nbsp;<math style="vertical-align: -3px">\vec{y}</math>&nbsp; is an eigenvector of &nbsp;<math style="vertical-align: 0px">A^{-1}.</math>&nbsp; What is the corresponding eigenvalue?
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 74: / Line 73: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{A^{-1}(A\vec{y})} & = & \displaystyle{A^{-1}(3\vec{y}}\\
+\displaystyle{A^{-1}(A\vec{y})} & = & \displaystyle{A^{-1}(3\vec{y})}\\
 &&\\
 & = & \displaystyle{3(A^{-1}\vec{y}).}
@@ Line 105: / Line 104: @@
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; See solution above.
 |}
-[[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]
+[[031_Review_Part_3|'''<u>Return to Review Problems</u>''']]

@@ Line 21: / Line 21: @@
 !Step 1: &nbsp;
 |-
-|Since &nbsp;<math>\vec{x}</math>&nbsp; is an eigenvector of &nbsp;<math>A</math>&nbsp; corresponding to the eigenvalue &nbsp;<math>2,</math>&nbsp; we know &nbsp;<math>\vec{x}\neq \vec{0}</math>&nbsp; and
+|Since &nbsp;<math style="vertical-align: 0px">\vec{x}</math>&nbsp; is an eigenvector of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; corresponding to the eigenvalue &nbsp;<math style="vertical-align: -4px">2,</math>&nbsp; we know &nbsp;<math style="vertical-align: -5px">\vec{x}\neq \vec{0}</math>&nbsp; and
 |-
+|
@@ Line 50: / Line 50: @@
 \end{array}</math>
 |-
-|Hence, since &nbsp;<math>\vec{x}\ne \vec{0},</math>&nbsp; we conclude that &nbsp;<math>\vec{x}</math>&nbsp; is an eigenvector of &nbsp;<math>A^3-A^2+I</math>&nbsp; corresponding to the eigenvalue &nbsp;<math>5.</math>
+|Hence, since &nbsp;<math style="vertical-align: -5px">\vec{x}\ne \vec{0},</math>&nbsp; we conclude that &nbsp;<math style="vertical-align: 0px">\vec{x}</math>&nbsp; is an eigenvector of &nbsp;<math style="vertical-align: -2px">A^3-A^2+I</math>&nbsp; corresponding to the eigenvalue &nbsp;<math style="vertical-align: 0px">5.</math>
 |}
@@ Line 59: / Line 59: @@
 !Step 1: &nbsp;
 |-
-|Since &nbsp;<math>\vec{y}</math>&nbsp; is an eigenvector of &nbsp;<math>A</math>&nbsp; corresponding to the eigenvalue &nbsp;<math>3,</math>&nbsp; we know &nbsp;<math>\vec{y}\neq \vec{0}</math>&nbsp; and
+|Since &nbsp;<math style="vertical-align: -3px">\vec{y}</math>&nbsp; is an eigenvector of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; corresponding to the eigenvalue &nbsp;<math style="vertical-align: -4px">3,</math>&nbsp; we know &nbsp;<math style="vertical-align: -5px">\vec{y}\neq \vec{0}</math>&nbsp; and
 |-
+|
 ::<math>A\vec{y}=3\vec{y}.</math>
 |-
-|Also, since &nbsp;<math>A</math>&nbsp; is invertible, &nbsp;<math>A^{-1}</math>&nbsp; exists.
+|Also, since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible, &nbsp;<math style="vertical-align: 0px">A^{-1}</math>&nbsp; exists.
 |}
@@ Line 70: / Line 70: @@
 !Step 2: &nbsp;
 |-
-|Now, we multiply the equation from Step 1 on the left by &nbsp;<math>A^{-1}</math> to obtain
+|Now, we multiply the equation from Step 1 on the left by &nbsp;<math style="vertical-align: 0px">A^{-1}</math>&nbsp; to obtain
 |-
+|
@@ Line 92: / Line 92: @@
 \end{array}</math>
 |-
-|Hence, &nbsp;<math>A^{-1}\vec{y}=\frac{1}{3}\vec{y}.</math>
+|Hence, &nbsp;<math style="vertical-align: -13px">A^{-1}\vec{y}=\frac{1}{3}\vec{y}.</math>
 |-
-|Therefore, &nbsp;<math>\vec{y}</math>&nbsp; is an eigenvector of &nbsp;<math>A^{-1}</math> corresponding to the eigenvalue &nbsp;<math>\frac{1}{3}.</math>
+|Therefore, &nbsp;<math style="vertical-align: -3px">\vec{y}</math>&nbsp; is an eigenvector of &nbsp;<math style="vertical-align: 0px">A^{-1}</math>&nbsp; corresponding to the eigenvalue &nbsp;<math style="vertical-align: -12px">\frac{1}{3}.</math>
 |}

@@ Line 20: / Line 20: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
 |-
+|
 |}
@@ Line 27: / Line 30: @@
 !Step 2: &nbsp;
 |-
+|Now, we have
 |}
@@ Line 34: / Line 58: @@
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 !Step 1: &nbsp;
 |-
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |}
@@ Line 48: / Line 101: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See solution above.
 |-
-|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; See solution above.
 |}
 [[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 02:25, 10 October 2017
Line 1:		Line 1:
−	<span class="exam">~~Consider the matrix~~  <math style="vertical-align: ~~-31px~~">A=	+	<span class="exam"> (a) Show that if  <math style="vertical-align: 0px">\vec{x}</math>  is an eigenvector of the matrix  <math style="vertical-align: 0px">A</math>  corresponding to the eigenvalue 2, then  <math style="vertical-align: 0px">\vec{x}</math>  is an eigenvector of  <math style="vertical-align: -2px">A^3-A^2+I.</math>  What is the corresponding eigenvalue?
−	\~~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=~~	+	<span class="exam">(b) Show that if  <math style="vertical-align: -3px">\vec{y}</math>  is an eigenvector of the matrix  <math style="vertical-align: 0px">A</math>  corresponding to the eigenvalue 3 and  <math style="vertical-align: 0px">A</math>  is invertible, then  <math style="vertical-align: -3px">\vec{y}</math>  is an eigenvector of  <math style="vertical-align: 0px">A^{-1}.</math>  What is the corresponding eigenvalue?
−	~~\begin{bmatrix}~~
−	~~1 & 0 & -1 & 5 \\~~
−	~~0 & -2 & 5 & -6 \\~~
−	~~0 & 0 & 0 & 0~~
−	~~\end{bmatrix}.</math>~~
−
−	<span class="exam">(a) ~~List rank~~  <math style="vertical-align: ~~0px">A</math>  and  <math style="vertical~~-~~align: 0px~~">\~~text~~{~~dim Nul~~ }A.</math>
−
−	~~<span class="exam">(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>

@@ Line 6: / Line 6: @@
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 !Foundations: &nbsp;
 |-
+|
 |}