Difference between revisions of "031 Review Part 2, Problem 5"

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations:    
 
!Foundations:    
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|Recall:
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|'''1.''' If the matrix &nbsp;<math style="vertical-align: 0px">B</math>&nbsp; is identical to the matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; except the entries in one of the rows of &nbsp;<math style="vertical-align: 0px">B</math>&nbsp;
 
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|-
 
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:are each equal to the corresponding entries of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; multiplied by the same scalar &nbsp;<math style="vertical-align: -4px">c,</math>&nbsp; then
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::<math>\text{det }B=c(\text{det }A).</math> 
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|'''2.''' &nbsp;<math style="vertical-align: -5px">\text{det } (AB)=(\text{det }A)(\text{det }B)</math>
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|'''3.''' For an invertible matrix &nbsp;<math style="vertical-align: -4px">A,</math>&nbsp; since &nbsp;<math style="vertical-align: 0px">AA^{-1}=I</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">\text{det }I=1,</math>&nbsp; we have
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::<math>\text{det }A^{-1}=\frac{1}{\text{det } A}.</math>
 
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Revision as of 15:35, 11 October 2017

Let    and    be    matrices with    and    Use properties of determinants to compute:

(a)  

(b)  


Foundations:  
Recall:
1. If the matrix    is identical to the matrix    except the entries in one of the rows of   
are each equal to the corresponding entries of    multiplied by the same scalar    then
2.  
3. For an invertible matrix    since    and    we have


Solution:

(a)

Step 1:  
Step 2:  

(b)

Step 1:  
Step 2:  


Final Answer:  
   (a)    
   (b)    

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