Rank of a matrix and more

1. Sep 6, 2007

rock.freak667

1. The problem statement, all variables and given/known data
Find the rank of the matrix A,where
$$A= \left( \begin{array}{cccc} 1 & 1 & 2 & 3\\ 4 & 3 & 5 & 16\\ 6 & 6 & 13 & 13\\ 14 & 12 & 23 & 45 \end{array} \right)$$

Find vectors$$x_0$$and$$e$$ such that any solution of the equation

$$Ax= \left( \begin{array}{c} 0\\ 2\\ -1\\ 3 \end{array} \right)$$ $$(*)$$
can be expressed in the form $$x_0+\lambdae$$ where $$\lambda\epsilonR$$

Hence show that there is no vector which satisfies $$*$$ and has all its elements positive

2. Relevant equations

First attempt at such a question, so unknown are any relevant equations

3. The attempt at a solution
Well for the first part to get the rank I put A in RRE form and then counted the number of non-zero rows and got for so $$r(A)=4$$

now for the second part,I thought to solve the equation by multiplying by $$A^{-1}$$ and finding $$x$$ but then I realized that I have no idea where to get $$x_0$$ or $$\lambda$$ or $$e$$

can anyone show me how to do these types of questions or can show me some similar example?

2. Sep 6, 2007

Hurkyl

Staff Emeritus
Well, you made a mistake somewhere in here.

You might have guessed that -- if you can write any solution in the form the problem asks for, what does the rank of the matrix have to be?

(Hint: what does the nullity of the matrix have to be?)

3. Sep 6, 2007

rock.freak667

Did I do the row-reduction wrong?
well from wikipedia...$$rank(A)+Nullity(A)=n$$ well $$n=4$$ in this case

BTW...This is the first time I have heard of nullity

4. Sep 6, 2007

Hurkyl

Staff Emeritus
I believe so. The statement of the problem implies the rank is not 4. (In fact, it implies a specific number for the rank) I tried once to do the row reduction myself, and I got the number I expected.

5. Sep 6, 2007

rock.freak667

Well I believe I did it over correctly and got $$r(A)=3$$

6. Sep 6, 2007

rootX

yes, you seems to be correct, if this is what you were trying to get:
$$\pmatrix{1 & 1 & 2 & 3\cr 0 & 1 & 3 & -4\cr 0 & 0 & 1 & -5\cr 0 & 0 & 0 & 0}$$

use maxima!!

http://aycu21.webshots.com/image/27020/2000682090404007350_rs.jpg

Last edited: Sep 6, 2007
7. Sep 6, 2007

rock.freak667

But how do I use the fact that $$r(A)=3$$ and the nullity to find the vectors in that form?

8. Sep 6, 2007

Hurkyl

Staff Emeritus
Well, how do you normally solve systems of equations? Have you tried that?

9. Sep 6, 2007

rock.freak667

Well normally for that matrix I would just augment it and try to put it in RRE form but then i dont know where $$x_0$$ and $$e$$ and $$\lambda$$ comes in

10. Sep 6, 2007

Hurkyl

Staff Emeritus
Well, try solving it first, then think about it.

By the way, you can edit your original post to fix that one formula; you're supposed to put spaces between things. And it looks a lot nicer if you use [ itex ] instead of [ tex ] for stuff in paragraphs.