Difference between revisions of "031 Review Part 1, Problem 6"

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(Created page with "<span class="exam">True or false: If all the entries of a  <math style="vertical-align: 0px">7\times 7</math>  matrix  <math style="vertical-align: 0px">A</math...")
 
 
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<span class="exam">True or false: If all the entries of a &nbsp;<math style="vertical-align: 0px">7\times 7</math>&nbsp; matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; are &nbsp;<math style="vertical-align: -4px">7,</math>&nbsp; then &nbsp;<math style="vertical-align: 0px">\text{det }A</math>&nbsp; must be &nbsp;<math style="vertical-align: 0px">7^7.</math>
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<span class="exam"> True or false: If &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a &nbsp;<math style="vertical-align: 0px">3\times 5</math>&nbsp; matrix and &nbsp;<math style="vertical-align: -4px">\text{dim Nul }A=2,</math>&nbsp; then &nbsp;<math style="vertical-align: 0px">A\vec{x}=\vec{b}</math>&nbsp; is consistent for all &nbsp;<math style="vertical-align: 0px">\vec{b}</math>&nbsp; in &nbsp;<math style="vertical-align: 0px">\mathbb{R}^3.</math>
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;  
 
!Solution: &nbsp;  
 
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|-
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|First, we switch to the limit to <math style="vertical-align: 0px">x</math> so that we can use L'Hopital's rule.
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|By the Rank Theorem, we have  
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|-
 
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|So, we have
 
 
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|-
 
|
 
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
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\displaystyle{\lim_{x \rightarrow \infty}\frac{3-2x^2}{5x^2 + x +1}} & \overset{L'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{-4x}{10x+1}}\\
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\displaystyle{5} & = & \displaystyle{\text{dim Col }A+\text{dim Nul }A}\\
 
&&\\
 
&&\\
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& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
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& = & \displaystyle{\text{dim Col }A+2.}
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&&\\
 
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& = & \displaystyle{-\frac{2}{5}}.
 
 
\end{array}</math>
 
\end{array}</math>
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|Hence, &nbsp;<math style="vertical-align: -2px">\text{dim Col }A=3.</math>
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|This tells us that &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; has three pivots.
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|Since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a &nbsp;<math style="vertical-align: 0px">3\times 5</math>&nbsp; matrix,
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|-
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|&nbsp;<math style="vertical-align: 0px">A</math>&nbsp; has a pivot in every row.
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|-
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|Therefore, &nbsp;<math style="vertical-align: 0px">A\vec{x}=\vec{b}</math>&nbsp; is consistent for all &nbsp;<math style="vertical-align: 0px">\vec{b}</math>&nbsp; in &nbsp;<math style="vertical-align: 0px">\mathbb{R}^3.</math>
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|-
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|So, the statement is true.
 
|}
 
|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|-
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|&nbsp;&nbsp; &nbsp; &nbsp; False
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|&nbsp;&nbsp; &nbsp; &nbsp; TRUE
 
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|}
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[[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]
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[[031_Review_Part_1|'''<u>Return to Review Problems</u>''']]

Latest revision as of 12:19, 15 October 2017

True or false: If    is a    matrix and    then    is consistent for all    in  

Solution:  
By the Rank Theorem, we have

       

Hence,  
This tells us that    has three pivots.
Since    is a    matrix,
   has a pivot in every row.
Therefore,    is consistent for all    in  
So, the statement is true.


Final Answer:  
       TRUE

Return to Review Problems

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