Difference between revisions of "031 Review Part 2, Problem 9"

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<span class="exam">If &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is an &nbsp;<math style="vertical-align: 0px">n\times n</math>&nbsp; matrix such that &nbsp;<math style="vertical-align: -4px">AA^T=I,</math>&nbsp; what are the possible values of &nbsp;<math style="vertical-align: 0px">\text{det }A?</math>
 
<span class="exam">If &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is an &nbsp;<math style="vertical-align: 0px">n\times n</math>&nbsp; matrix such that &nbsp;<math style="vertical-align: -4px">AA^T=I,</math>&nbsp; what are the possible values of &nbsp;<math style="vertical-align: 0px">\text{det }A?</math>
 
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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|'''1.''' &nbsp;<math style="vertical-align: -5px">\text{det }(AB)=(\text{det }A)(\text{det }B)</math>
 
|'''1.''' &nbsp;<math style="vertical-align: -5px">\text{det }(AB)=(\text{det }A)(\text{det }B)</math>
 
|-
 
|-
|'''2.''' &nbsp;<math style="vertical-align: 0px">\text{det }I=n</math>
+
|'''2.''' &nbsp;<math style="vertical-align: 0px">\text{det }I=1</math>
 
|-
 
|-
 
|'''3.''' &nbsp;<math style="vertical-align: 0px">\text{det }A^T=\text{det }A</math>
 
|'''3.''' &nbsp;<math style="vertical-align: 0px">\text{det }A^T=\text{det }A</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 +
|-
 +
|Using the facts in the Foundations section, we have
 
|-
 
|-
 
|
 
|
 +
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 +
\displaystyle{1} & = & \displaystyle{\text{det }(I)}\\
 +
&&\\
 +
& = & \displaystyle{\text{det } (AA^T)}\\
 +
&&\\
 +
& = & \displaystyle{(\text{det }A)(\text{det } A^T)}\\
 +
&&\\
 +
& = & \displaystyle{(\text{det }A)(\text{det } A)}\\
 +
&&\\
 +
& = & \displaystyle{(\text{det }A)^2.}
 +
\end{array}</math>
 
|}
 
|}
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 +
|-
 +
|Taking the square root of both sides of the equation
 
|-
 
|-
 
|
 
|
 +
::<math>(\text{det }A)^2=1,</math>
 +
|-
 +
|we obtain &nbsp;<math style="vertical-align: -1px">\text{det }A=\pm 1.</math>
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|&nbsp;&nbsp; &nbsp; &nbsp;  
+
|&nbsp;&nbsp; &nbsp; &nbsp; <math>\text{det }A=\pm 1</math>
 
|}
 
|}
[[031_Review_Part_2|'''<u>Return to Sample Exam</u>''']]
+
[[031_Review_Part_2|'''<u>Return to Review Problems</u>''']]

Latest revision as of 13:40, 15 October 2017

If    is an    matrix such that    what are the possible values of  

Foundations:  
Recall:
1.  
2.  
3.  


Solution:

Step 1:  
Using the facts in the Foundations section, we have

       

Step 2:  
Taking the square root of both sides of the equation
we obtain  


Final Answer:  
      

Return to Review Problems