Difference between revisions of "031 Review Part 2, Problem 9"
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Kayla Murray (talk | contribs) |
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 1: | !Step 1: | ||
+ | |- | ||
+ | |Using the facts in the Foundations section, we have | ||
|- | |- | ||
| | | | ||
+ | <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: | !Step 2: | ||
+ | |- | ||
+ | |Taking the square root of both sides of the equation | ||
|- | |- | ||
| | | | ||
+ | ::<math>(\text{det }A)^2=1,</math> | ||
+ | |- | ||
+ | |we obtain <math style="vertical-align: -1px">\text{det }A=\pm 1.</math> | ||
|} | |} | ||
Line 33: | Line 51: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
− | | | + | | <math>\text{det }A=\pm 1</math> |
|} | |} | ||
[[031_Review_Part_2|'''<u>Return to Sample Exam</u>''']] | [[031_Review_Part_2|'''<u>Return to Sample Exam</u>''']] |
Revision as of 19:30, 10 October 2017
If is an matrix such that what are the possible values of
Foundations: |
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Recall: |
1. |
2. |
3. |
Solution:
Step 1: |
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Using the facts in the Foundations section, we have |
|
Step 2: |
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Taking the square root of both sides of the equation |
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we obtain |
Final Answer: |
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