Difference between revisions of "031 Review Part 2, Problem 1"
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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: | ||
+ | |- | ||
+ | |From the matrix <math style="vertical-align: -4px">B,</math> we see that <math style="vertical-align: 0px">A</math> contains two pivots. | ||
+ | |- | ||
+ | |Therefore, | ||
|- | |- | ||
| | | | ||
+ | <math>\begin{array}{rcl} | ||
+ | \displaystyle{\text{rank }A} & = & \displaystyle{\text{dim Col }A}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{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: | ||
+ | |- | ||
+ | |By the Rank Theorem, we have | ||
|- | |- | ||
| | | | ||
+ | <math>\begin{array}{rcl} | ||
+ | \displaystyle{4} & = & \displaystyle{\text{rank }A+\text{dim Nul }A}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{2+\text{dim Nul }A.} | ||
+ | \end{array}</math> | ||
+ | |- | ||
+ | |Hence, <math style="vertical-align: -2px">\text{dim Nul }A=2.</math> | ||
|} | |} | ||
Revision as of 10:22, 10 October 2017
Consider the matrix and assume that it is row equivalent to the matrix
(a) List rank and
(b) Find bases for and Find an example of a nonzero vector that belongs to as well as an example of a nonzero vector that belongs to
Foundations: |
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1. For a matrix the rank of is |
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2. is the vector space spanned by the columns of |
3. is the vector space containing all solutions to |
Solution:
(a)
Step 1: |
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From the matrix we see that contains two pivots. |
Therefore, |
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Step 2: |
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By the Rank Theorem, we have |
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Hence, |
(b)
Step 1: |
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Step 2: |
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Final Answer: |
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(a) |
(b) |