Difference between revisions of "031 Review Part 1, Problem 4"

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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=   
 +
    \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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|-
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|So, we have
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|First, notice that &nbsp;<math style="vertical-align: -5px">\text{det }A=1\neq 0.</math>
 
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|-
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|
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|Therefore, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible.
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
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|-
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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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|Since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a diagonal matrix, the eigenvalues of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; are the entries on the diagonal.
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&&\\
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|-
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& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
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|Therefore, the only eigenvalue of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">1.</math>&nbsp; Additionally, there is only one linearly independent eigenvector.
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&&\\
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|-
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& = & \displaystyle{-\frac{2}{5}}.
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|Hence, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is not diagonalizable and the statement is false.
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\end{array}</math>
 
 
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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; False
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|&nbsp;&nbsp; &nbsp; &nbsp; FALSE
 
|}
 
|}
 
[[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]
 
[[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 13:43, 9 October 2017

True or false: If    is invertible, then    is diagonalizable.

Solution:  
Let  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= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}.}  
First, notice 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 \text{det }A=1\neq 0.}
Therefore,  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.
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 A}   is a diagonal matrix, the eigenvalues 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}   are the entries on the diagonal.
Therefore, the only eigenvalue 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}   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 1.}   Additionally, there is only one linearly independent eigenvector.
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 A}   is not diagonalizable and the statement is false.
Final Answer:  
       FALSE

Return to Sample Exam

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