Difference between revisions of "031 Review Part 2, Problem 10"

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(Created page with "<span class="exam">Consider the matrix  <math style="vertical-align: -31px">A= \begin{bmatrix} 1 & -4 & 9 & -7 \\ -1 & 2 & -4 & 1 \\...")
 
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<span class="exam">Consider the matrix &nbsp;<math style="vertical-align: -31px">A=  
+
<span class="exam">(a) Suppose a &nbsp;<math style="vertical-align: 0px">6\times 8</math>&nbsp; matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; has 4 pivot columns. What is &nbsp;<math style="vertical-align: -1px">\text{dim Nul }A?</math>&nbsp; Is &nbsp;<math style="vertical-align: -1px">\text{Col }A=\mathbb{R}^4?</math>&nbsp; Why or why not?
    \begin{bmatrix}
 
          1 & -4 & 9 & -7 \\
 
          -1 & 2  & -4 & 1 \\
 
          5 & -6 & 10 & 7
 
        \end{bmatrix}</math>&nbsp; and assume that it is row equivalent to the matrix
 
  
::<math>B=   
+
<span class="exam">(b) If &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a &nbsp;<math style="vertical-align: 0px">7\times 5</math>&nbsp; matrix, what is the smallest possible dimension of &nbsp;<math style="vertical-align: -1px">\text{Nul }A?</math>
    \begin{bmatrix}
 
          1 & 0 & -1 & 5 \\
 
          0 & -2  & 5 & -6 \\
 
          0 & 0 & 0 & 0
 
        \end{bmatrix}.</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">(b) Find bases for &nbsp;<math style="vertical-align: 0px">\text{Col }A</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">\text{Nul }A.</math>&nbsp; Find an example of a nonzero vector that belongs to &nbsp;<math style="vertical-align: -5px">\text{Col }A,</math>&nbsp; as well as an example of a nonzero vector that belongs to &nbsp;<math style="vertical-align: 0px">\text{Nul }A.</math>
 
  
  

Revision as of 19:17, 9 October 2017

(a) Suppose a    matrix    has 4 pivot columns. What is    Is    Why or why not?

(b) If    is a    matrix, what is the smallest possible dimension of  


Foundations:  


Solution:

(a)

Step 1:  
Step 2:  

(b)

Step 1:  
Step 2:  


Final Answer:  
   (a)    
   (b)    

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