Difference between revisions of "031 Review Part 2, Problem 1"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
| Line 38: | Line 38: | ||
{| 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: |
|---|
| 1. For a matrix the rank of is |
|
|
| 2. is the vector space spanned by the columns of |
| 3. is the vector space containing all solutions to |
Solution:
(a)
| Step 1: |
|---|
| From the matrix we see that contains two pivots. |
| Therefore, |
|
|
| Step 2: |
|---|
| By the Rank Theorem, we have |
|
|
| Hence, |
(b)
| Step 1: |
|---|
| Step 2: |
|---|
| Final Answer: |
|---|
| (a) |
| (b) |