Difference between revisions of "031 Review Part 2, Problem 5"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
Line 8: | Line 8: | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Foundations: | !Foundations: | ||
+ | |- | ||
+ | |Recall: | ||
+ | |- | ||
+ | |'''1.''' If the matrix <math style="vertical-align: 0px">B</math> is identical to the matrix <math style="vertical-align: 0px">A</math> except the entries in one of the rows of <math style="vertical-align: 0px">B</math> | ||
|- | |- | ||
| | | | ||
+ | :are each equal to the corresponding entries of <math style="vertical-align: 0px">A</math> multiplied by the same scalar <math style="vertical-align: -4px">c,</math> then | ||
+ | |- | ||
+ | | | ||
+ | ::<math>\text{det }B=c(\text{det }A).</math> | ||
+ | |- | ||
+ | |'''2.''' <math style="vertical-align: -5px">\text{det } (AB)=(\text{det }A)(\text{det }B)</math> | ||
+ | |- | ||
+ | |'''3.''' For an invertible matrix <math style="vertical-align: -4px">A,</math> since <math style="vertical-align: 0px">AA^{-1}=I</math> and <math style="vertical-align: -4px">\text{det }I=1,</math> we have | ||
+ | |- | ||
+ | | | ||
+ | ::<math>\text{det }A^{-1}=\frac{1}{\text{det } A}.</math> | ||
|} | |} | ||
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 |
|
|
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) |