Difference between revisions of "031 Review Part 2, Problem 1"

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1:    
 
!Step 1:    
 +
|-
 +
|From the matrix &nbsp;<math style="vertical-align: -4px">B,</math>&nbsp; we see that &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; contains two pivots.
 +
|-
 +
|Therefore,
 
|-
 
|-
 
|
 
|
 +
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 +
\displaystyle{\text{rank }A} & = & \displaystyle{\text{dim Col }A}\\
 +
&&\\
 +
& = & \displaystyle{2.}
 +
\end{array}</math>
 
|}
 
|}
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 +
|-
 +
|By the Rank Theorem, we have
 
|-
 
|-
 
|
 
|
 +
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 +
\displaystyle{4} & = & \displaystyle{\text{rank }A+\text{dim Nul }A}\\
 +
&&\\
 +
& = & \displaystyle{2+\text{dim Nul }A.}
 +
\end{array}</math>
 +
|-
 +
|Hence, &nbsp;<math style="vertical-align: -2px">\text{dim Nul }A=2.</math>
 
|}
 
|}
  

Revision as of 10:22, 10 October 2017

Consider the matrix    and assume that it is row equivalent to the matrix

(a) List rank    and  

(b) Find bases for    and    Find an example of a nonzero vector that belongs to    as well as an example of a nonzero vector that belongs to  


Foundations:  
1. For a matrix    the rank of    is
2.    is the vector space spanned by the columns of  
3.    is the vector space containing all solutions to  


Solution:

(a)

Step 1:  
From the matrix    we see that    contains two pivots.
Therefore,

       

Step 2:  
By the Rank Theorem, we have

       

Hence,  

(b)

Step 1:  
Step 2:  


Final Answer:  
   (a)    
   (b)    

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