Difference between revisions of "031 Review Part 2, Problem 3"

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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">Let 
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&nbsp;<math>B=     
 
     \begin{bmatrix}
 
     \begin{bmatrix}
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           1 & -4 & 9 & -7 \\
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           1 & -2 & 3 & 4\\
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          -1 & 2  & -4 & 1 \\
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           0 & 3 &0 &0\\
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           5 & -6 & 10 & 7
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           0 & 5 & 1 & 2\\
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        \end{bmatrix}</math>&nbsp; and assume that it is row equivalent to the matrix
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           0 & -1 & 3 & 6
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         \end{bmatrix}.
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::<math>B=   
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</math>
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    \begin{bmatrix}
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           1 & 0 & -1 & 5 \\
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<span class="exam">(a) Is &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; invertible? Explain.
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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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<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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 +
|-
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|'''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>
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|-
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|'''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
 
|-
 
|-
 
|
 
|
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::if and only if the columns of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; span &nbsp;<math style="vertical-align: 0px">\mathbb{R}^m.</math>
 
|}
 
|}
  
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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 calculating &nbsp;<math style="vertical-align: -1px">\text{det }B.</math>
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|-
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|To do this, we use cofactor expansion along the second row first and then the first column.
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|-
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|So, we have
 
|-
 
|-
 
|
 
|
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
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\displaystyle{\text{det }B} & = & \displaystyle{3(-1)^{2+2}\left|\begin{array}{ccc} 
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          1 & 3 & 4 \\
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          0 & 1  & 2 \\
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          0 & 3 & 6
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        \end{array}\right|}\\
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&&\\
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& = & \displaystyle{3\cdot 1 \cdot (-1)^{1+1} \left|\begin{array}{cc} 
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          1 & 2 \\
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          3 & 6
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        \end{array}\right|}\\
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&&\\
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& = & \displaystyle{3(6-6)}\\
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&&\\
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& = & \displaystyle{0.}
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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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|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.
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|
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|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>
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|-
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|This would mean that &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; contains 4 pivots.
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|
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|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.
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|-
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|Hence, &nbsp;<math style="vertical-align: 0px">T</math>&nbsp; is not onto.
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
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|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;  
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|&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.
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|-
 
|-
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|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;  
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|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; No, see explaination above.
 
|}
 
|}
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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 13:17, 15 October 2017

Let  

(a) Is    invertible? Explain.

(b) Define a linear transformation    by the formula  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T(\vec{x})=B\vec{x}.}   Is  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T}   onto? Explain.

Foundations:  
1. A matrix  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A}   is invertible if and only if  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{det }A\neq 0.}
2. A linear transformation  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T}   given by  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T(\vec{x})=A\vec{x},}   where  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A}   is a  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\times n}   matrix, is onto
if and only if the columns of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A}   span  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}^m.}


Solution:

(a)

Step 1:  
We begin by calculating  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{det }B.}
To do this, we use cofactor expansion along the second row first and then the first column.
So, we have

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\text{det }B} & = & \displaystyle{3(-1)^{2+2}\left|\begin{array}{ccc} 1 & 3 & 4 \\ 0 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right|}\\ &&\\ & = & \displaystyle{3\cdot 1 \cdot (-1)^{1+1} \left|\begin{array}{cc} 1 & 2 \\ 3 & 6 \end{array}\right|}\\ &&\\ & = & \displaystyle{3(6-6)}\\ &&\\ & = & \displaystyle{0.} \end{array}}

Step 2:  
Since  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{det }B=0,}   we know that  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B}   is not invertible.

(b)

Step 1:  
If  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T}   was onto, then  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B}   spans  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}^4.}
This would mean that  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B}   contains 4 pivots.
Step 2:  
But, if  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B}   has 4 pivots, then  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B}   would be invertible, which is not true.
Hence,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T}   is not onto.


Final Answer:  
   (a)     Since  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{det }B=0,}   we have that  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B}   is not invertible.
   (b)     No, see explaination above.

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