Difference between revisions of "031 Review Part 1, Problem 7"

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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: Let &nbsp;<math style="vertical-align: 0px">C=AB</math>&nbsp; for &nbsp;<math style="vertical-align: 0px">4\times 4</math>&nbsp; matrices &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">B.</math>&nbsp; If &nbsp;<math style="vertical-align: 0px">C</math>&nbsp; is invertible, then &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible.
  
 
{| 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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|If &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is not invertible, then &nbsp;<math style="vertical-align: 0px">\text{det } A=0.</math>
 
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|-
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|So, we have
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|Since &nbsp;<math style="vertical-align: -5px">C=AB,</math>&nbsp; we have
 
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
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::<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{\text{det}C} & = & \displaystyle{\text{det} (AB)}\\
 
&&\\
 
&&\\
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& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
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& = & \displaystyle{(\text{det} A) \cdot (\text{det} B)}\\
 
&&\\
 
&&\\
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& = & \displaystyle{-\frac{2}{5}}.
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& = & \displaystyle{0\cdot (\text{det} B)}\\
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\end{array}</math>
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&&\\
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& = & \displaystyle{0.}
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\end{array}</math>  
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|Since &nbsp;<math style="vertical-align: -5px">\text{det } C =0,</math>&nbsp; we know &nbsp;<math style="vertical-align: 0px">C</math>&nbsp; is not invertible, which is a contradiction.
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|So, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; must be invertible and 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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|&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 12:20, 15 October 2017

True or false: Let    for    matrices    and    If    is invertible, then    is invertible.

Solution:  
If    is not invertible, then  
Since    we have
Since    we know    is not invertible, which is a contradiction.
So,    must be invertible and the statement is true.


Final Answer:  
       TRUE

Return to Review Problems

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