Difference between revisions of "031 Review Part 1, Problem 5"

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!Solution:    
 
!Solution:    
 
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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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|Let &nbsp;<math style="vertical-align: -20px">A=   
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    \begin{bmatrix}
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          1 & 1  \\
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          0 & 1
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        \end{bmatrix}</math>&nbsp; and &nbsp;<math style="vertical-align: -20px">B=   
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    \begin{bmatrix}
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          -1 & -1 \\
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          0 & -1
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        \end{bmatrix}.</math>&nbsp;
 
|-
 
|-
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|So, we have
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|Notice that
 
|-
 
|-
 
|
 
|
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
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::<math>\text{det} A=\text{det} B=1.</math>
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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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|-
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&&\\
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|Since &nbsp;<math style="vertical-align: -6px">\text{det} A,\text{det} B\neq 0,</math>&nbsp; we know &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; are invertible.
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& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
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|-
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&&\\
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|But,
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& = & \displaystyle{-\frac{2}{5}}.
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|-
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\end{array}</math>
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|
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::<math style="vertical-align: -31px">A+B=   
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    \begin{bmatrix}
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          0 & 0 \\
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          0 & 0
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        \end{bmatrix},</math>&nbsp;
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|-
 +
|which is not invertible.
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|-
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|Hence, this statement is false.
 
|}
 
|}
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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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|&nbsp;&nbsp; &nbsp; &nbsp; False
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|&nbsp;&nbsp; &nbsp; &nbsp; FALSE
 
|}
 
|}
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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 12:18, 15 October 2017

True or false: If    and    are invertible    matrices, then so is  

Solution:  
Let    and   
Notice that
Since    we know    and    are invertible.
But,
 
which is not invertible.
Hence, this statement is false.


Final Answer:  
       FALSE

Return to Review Problems

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