Difference between revisions of "031 Review Part 1, Problem 9"

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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 an invertible &nbsp;<math style="vertical-align: 0px">3\times 3</math>&nbsp; matrix, and &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">C</math>&nbsp; are &nbsp;<math style="vertical-align: 0px">3\times 3</math>&nbsp; matrices such that &nbsp;<math style="vertical-align: -4px">AB=AC,</math>&nbsp;  
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<span class="exam">then &nbsp;<math style="vertical-align: 0px">B=C.</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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|Since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible, &nbsp;<math style="vertical-align: 0px">A^{-1}</math>&nbsp; exists.
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|-
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|Since &nbsp;<math style="vertical-align: -4px">AB=AC,</math>&nbsp; we have
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|
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::<math>A^{-1}(AB)=A^{-1}(AC).</math>
 
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|-
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|So, we have
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|Then, by associativity of matrix multiplication, 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{(A^{-1}A)B} & = & \displaystyle{(A^{-1}A)C}\\
 
&&\\
 
&&\\
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& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
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\displaystyle{I_3 B} & = & \displaystyle{I_3 C}\\
 
&&\\
 
&&\\
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& = & \displaystyle{-\frac{2}{5}}.
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\displaystyle{B} & = & \displaystyle{C}
 
\end{array}</math>
 
\end{array}</math>
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|-
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|where &nbsp;<math style="vertical-align: -3px">I_3</math>&nbsp; is the &nbsp;<math style="vertical-align: -1px">3\times 3</math>&nbsp; identity matrix.
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|-
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|Hence, 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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[[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 12:22, 15 October 2017

True or false: If    is an invertible    matrix, and    and    are    matrices such that   

then  

Solution:  
Since    is invertible,    exists.
Since    we have
Then, by associativity of matrix multiplication, we have

       

where    is the    identity matrix.
Hence, the statement is true.


Final Answer:  
       TRUE

Return to Review Problems

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