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

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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=   
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<span class="exam"> Find the dimension of the subspace spanned by the given vectors. Are these vectors linearly independent?
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    \begin{bmatrix}
 
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          1 & -4 & 9 & -7 \\
 
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          -1 & 2  & -4 & 1 \\
 
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          5 & -6 & 10 & 7
 
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        \end{bmatrix}</math>&nbsp; and assume that it is row equivalent to the matrix
 
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::<math>B=   
 
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    \begin{bmatrix}
 
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          1 & 0 & -1 & 5 \\
 
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          0 & -2  & 5 & -6 \\
 
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          0 & 0 & 0 & 0
 
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        \end{bmatrix}.</math>     
 
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<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>
 
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<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>
 
  
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::<math>\begin{bmatrix}
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          1  \\
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          0 \\
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          2
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        \end{bmatrix},
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        \begin{bmatrix}
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          3  \\
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          1 \\
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          1
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        \end{bmatrix},
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        \begin{bmatrix}
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          -2  \\
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          -1 \\
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          1
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        \end{bmatrix},
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        \begin{bmatrix}
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          5  \\
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          2 \\
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          2
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        \end{bmatrix}</math>
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
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|-
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|'''1.''' &nbsp;<math style="vertical-align: -1px">\text{dim Col }A</math>&nbsp; is the number of pivots in &nbsp;<math style="vertical-align: 0px">A.</math>
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|-
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|'''2.''' A set of vectors &nbsp;<math style="vertical-align: -4px">\{\vec{v_1},\vec{v_2},\ldots,\vec{v_n}\}</math>&nbsp; is linearly independent if
 
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|-
 
|
 
|
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::the only solution to &nbsp;<math style="vertical-align: -4px">x_1\vec{v_1}+x_2\vec{v_2}+\cdots+x_n\vec{v_n}=\vec{0}</math>&nbsp; is the trivial solution.
 
|}
 
|}
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
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|-
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|We begin by putting these vectors together in a matrix. So, we have
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|-
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|
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::<math>   
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    \begin{bmatrix}
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          1 & 3 & -2 & 5 \\
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          0 & 1  & -1 & 2 \\
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          2 & 1 & 1 & 2
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        \end{bmatrix}.</math>
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|-
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|Now, we row reduce this matrix. We get
 
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|-
 
|
 
|
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
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\displaystyle{\left[\begin{array}{cccc} 
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          1 & 3 & -2 & 5 \\
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          0 & 1  & -1 & 2 \\
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          2 & 1 & 1 & 2
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        \end{array}\right]} & \sim & \displaystyle{\left[\begin{array}{cccc} 
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          1 & 3 & -2 & 5 \\
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          0 & 1  & -1 & 2 \\
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          0 & -5 & 5 & -8
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        \end{array}\right]}\\
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&&\\
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& \sim & \displaystyle{\left[\begin{array}{cccc} 
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          1 & 3 & -2 & 5 \\
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          0 & 1  & -1 & 2 \\
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          0 & 0 & 0 & 2
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        \end{array}\right]}
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\end{array}</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|
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|Now, we have 3 pivots in this matrix. So, the dimension of the column space of the matrix we started with is 3.
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|-
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|Hence, the dimension of the subspace spanned by these vectors is &nbsp;<math style="vertical-align: 0px">3.</math>
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|-
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|When we row reduced the matrix, we had a column that did not contain a pivot.
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|-
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|This means we have a free variable in the system corresponding to &nbsp;<math style="vertical-align: 0px">Ax=0.</math>&nbsp;
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|-
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|So, these vectors are not linearly independent.
 
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|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|-
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|&nbsp;&nbsp; &nbsp; &nbsp;   
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|&nbsp;&nbsp; &nbsp; &nbsp;  The dimension is &nbsp;<math style="vertical-align: 0px">3</math>&nbsp; and the vectors are not linearly independent.
 
|}
 
|}
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[[031_Review_Part_2|'''<u>Return to Sample Exam</u>''']]
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[[031_Review_Part_2|'''<u>Return to Review Problems</u>''']]

Latest revision as of 13:13, 15 October 2017

Find the dimension of the subspace spanned by the given vectors. Are these vectors linearly independent?

Foundations:  
1.    is the number of pivots in  
2. A set of vectors    is linearly independent if
the only solution to    is the trivial solution.


Solution:

Step 1:  
We begin by putting these vectors together in a matrix. So, we have
Now, we row reduce this matrix. We get

       

Step 2:  
Now, we have 3 pivots in this matrix. So, the dimension of the column space of the matrix we started with is 3.
Hence, the dimension of the subspace spanned by these vectors is  
When we row reduced the matrix, we had a column that did not contain a pivot.
This means we have a free variable in the system corresponding to   
So, these vectors are not linearly independent.


Final Answer:  
       The dimension is    and the vectors are not linearly independent.

Return to Review Problems

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