Rank of a Matrix and Solving Linear Equations with Vectors

[rock.freak667]

Homework Statement

Find the rank of the matrix A,where
$$A= \left(<br /> \begin{array}{cccc}<br /> 1 & 1 & 2 & 3\\<br /> 4 & 3 & 5 & 16\\<br /> 6 & 6 & 13 & 13\\<br /> 14 & 12 & 23 & 45 <br /> \end{array}<br /> \right)$$

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

$$Ax= \left(<br /> \begin{array}{c}<br /> 0\\<br /> 2\\<br /> -1\\<br /> 3 <br /> \end{array}<br /> \right)$$ $$(*)$$
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

Homework Equations

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

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?

 

[Hurkyl]

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 $$r(A)=4$$

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?)

 

[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

 

[Hurkyl]

Did I do the row-reduction wrong?

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.

 

[rock.freak667]

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

 

[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

 

[rock.freak667]

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

 

[Hurkyl]

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

 

[rock.freak667]

Well normally for that matrix I would just augment it and try to put it in RRE form but then i don't know where $$x_0$$ and $$e$$ and $$\lambda$$ comes in

 

[Hurkyl]

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.