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

• Feb 27th 2011, 12:35 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 $\displaystyle $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?
• Feb 27th 2011, 01:19 PM
tonio
Quote:

Originally Posted by Jskid
Find a set of vectors spanning the solution space of Ax=0 where $\displaystyle $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
• Feb 27th 2011, 02:05 PM
Jskid
I did it wrong, $\displaystyle 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 $\displaystyle x_1=-r$
$\displaystyle x_2=-r$
$\displaystyle x_3=r$
$\displaystyle x_4$ I'm not sure what to do with
• Feb 27th 2011, 02:50 PM
Plato
Have you considered
$\displaystyle $\left[ {\begin{array}{r} 1 \\ 1 \\ { - 1} \\ 0 \\ \end{array} } \right]~? • Feb 27th 2011, 04:42 PM Jskid I didn't realize a row of 0s could be -r I thought it had to be r • Feb 28th 2011, 02:45 AM HallsofIvy Quote: Originally Posted by Jskid I did it wrong, \displaystyle 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 $\displaystyle x_1=-r$
$\displaystyle x_2=-r$
$\displaystyle x_3=r$
$\displaystyle x_4$ 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 $\displaystyle x_1+ x_3= 0$, $\displaystyle x_2+ x_3= 0$, and $\displaystyle x_4= 0$. The first two equations can be solved for $\displaystyle x_1= -x_3$, $\displaystyle x_2= -x_3$. Taking $\displaystyle x_3$ to be your "r", we have $\displaystyle x_1= -r$, $\displaystyle x_2= -r$, $\displaystyle x_3= r$ and, of course, $\displaystyle x_4= 0$. Taking r= -1 gives Plato's solution and all solutions are a multiple of that.