031 Review Part 2, Problem 3 - Revision history

Kayla Murray at 20:17, 15 October 2017

2017-10-15T20:17:19Z

Kayla Murray at 15:03, 11 October 2017

2017-10-11T15:03:52Z

Kayla Murray at 04:20, 11 October 2017

2017-10-11T04:20:09Z

Kayla Murray at 02:08, 10 October 2017

2017-10-10T02:08:22Z

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

2017-10-10T01:56:38Z

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)'''

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

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!Step 1:  
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!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)'''    
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[[031_Review_Part_2|'''Return to Sample Exam''']]

@@ Line 12: / Line 12: @@
 <span class="exam">(b) Define a linear transformation &nbsp;<math style="vertical-align: 0px">T</math>&nbsp; by the formula &nbsp;<math style="vertical-align: -5px">T(\vec{x})=B\vec{x}.</math>&nbsp; Is &nbsp;<math style="vertical-align: 0px">T</math>&nbsp; onto? Explain.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 19: / Line 18: @@
 |'''1.''' A matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible if and only if &nbsp;<math style="vertical-align: -5px">\text{det }A\neq 0.</math>
 |-
-|'''2.''' A linear transformation &nbsp;<math style="vertical-align: 0px">T</math>&nbsp; given by &nbsp;<math style="vertical-align: -5px">T(\vec{x})=A\vec{x}</math>&nbsp; where &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a &nbsp;<math style="vertical-align: 0px">m\times n</math>&nbsp; matrix
+|'''2.''' A linear transformation &nbsp;<math style="vertical-align: 0px">T</math>&nbsp; given by &nbsp;<math style="vertical-align: -5px">T(\vec{x})=A\vec{x},</math>&nbsp; where &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a &nbsp;<math style="vertical-align: 0px">m\times n</math>&nbsp; matrix, is onto
 |-
+|
@@ Line 61: / Line 60: @@
 !Step 2: &nbsp;
 |-
-|Since &nbsp;<math style="vertical-align: -4px">\text{det }B=0,</math>&nbsp; we have that &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; is not invertible.
+|Since &nbsp;<math style="vertical-align: -4px">\text{det }B=0,</math>&nbsp; we know that &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; is not invertible.
 |}
@@ Line 91: / Line 90: @@
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; No, see explaination above.
 |}
-[[031_Review_Part_2|'''<u>Return to Sample Exam</u>''']]
+[[031_Review_Part_2|'''<u>Return to Review Problems</u>''']]

@@ Line 32: / Line 32: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
 |-
+|
 |}
@@ Line 39: / Line 61: @@
 !Step 2: &nbsp;
 |-
+|Since &nbsp;<math style="vertical-align: -4px">\text{det }B=0,</math>&nbsp; we have that &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; is not invertible.
 |}
@@ Line 47: / Line 69: @@
 !Step 1: &nbsp;
 |-
+|If &nbsp;<math style="vertical-align: 0px">T</math>&nbsp; was onto, then &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; spans &nbsp;<math style="vertical-align: 0px">\mathbb{R}^4.</math>
 |}
@@ Line 53: / Line 77: @@
 !Step 2: &nbsp;
 |-
+|But, if &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; has 4 pivots, then &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; would be invertible, which is not true.
 |}
@@ Line 60: / Line 86: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; Since &nbsp;<math style="vertical-align: -4px">\text{det }B=0,</math>&nbsp; we have that &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; is not invertible.
 |-
-|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; No, see explaination above.
 |}
 [[031_Review_Part_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the matrix &nbsp;<math style="vertical-align: -31px">A=
+<span class="exam">Let
      \begin{bmatrix}
-& -4 & 9 & -7 \\
+& -2 & 3 & 4\\
-            -1 & 2  & -4 & 1 \\
+& 3 &0 &0\\
-& -6 & 10 & 7
+& 5 & 1 & 2\\
-          \end{bmatrix}</math>&nbsp; and assume that it is row equivalent to the matrix
+& -1 & 3 & 6
-::<math>B=
+<span class="exam">(b) Define a linear transformation &nbsp;<math style="vertical-align: 0px">T</math>&nbsp; by the formula &nbsp;<math style="vertical-align: -5px">T(\vec{x})=B\vec{x}.</math>&nbsp; Is &nbsp;<math style="vertical-align: 0px">T</math>&nbsp; onto? Explain.