Find a set of vectors spanning the solution space, stuck at last step

• February 27th 2011, 01:35 PM
Jskid
Find a set of vectors spanning the solution space, stuck at last step
Find a set of vectors spanning the solution space of Ax=0 where $
$A = \left[ {\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 1 & 2 & 3 & 1 \\ 2 & 1 & 3 & 1 \\ 1 & 1 & 2 & 1 \\ \end{array} } \right]$
$

So I rref and it's row equivalent to the identity matrix. How do I interpret the result?
• February 27th 2011, 02:19 PM
tonio
Quote:

Originally Posted by Jskid
Find a set of vectors spanning the solution space of Ax=0 where $
$A = \left[ {\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 1 & 2 & 3 & 1 \\ 2 & 1 & 3 & 1 \\ 1 & 1 & 2 & 1 \\ \end{array} } \right]$
$

So I rref and it's row equivalent to the identity matrix. How do I interpret the result?

If the matrix is row-equivalent to the unit matrix then the only solution to the adjoint homogeneous

system is the trivial one...

Tonio
• February 27th 2011, 03:05 PM
Jskid
I did it wrong, $rref(A) = $\left[ {\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} } \right]$
$

So $x_1=-r$
$x_2=-r$
$x_3=r$
$x_4$ I'm not sure what to do with
• February 27th 2011, 03:50 PM
Plato
Have you considered
$$\left[ {\begin{array}{r} 1 \\ 1 \\ { - 1} \\ 0 \\ \end{array} } \right]~?$ • February 27th 2011, 05:42 PM Jskid I didn't realize a row of 0s could be -r I thought it had to be r • February 28th 2011, 03:45 AM HallsofIvy Quote: Originally Posted by Jskid I did it wrong, $rref(A) = \[ \left[ {\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} } \right]$
$

So $x_1=-r$
$x_2=-r$
$x_3=r$
$x_4$ I'm not sure what to do with

I have no clue what "r" means here. Is it just an arbitrary number?

The equations corresponding to your reduced matrix are $x_1+ x_3= 0$, $x_2+ x_3= 0$, and $x_4= 0$. The first two equations can be solved for $x_1= -x_3$, $x_2= -x_3$. Taking $x_3$ to be your "r", we have $x_1= -r$, $x_2= -r$, $x_3= r$ and, of course, $x_4= 0$. Taking r= -1 gives Plato's solution and all solutions are a multiple of that.