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

**Jskid** · February 27th 2011, 12:35 PM

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?

**tonio** · February 27th 2011, 01:19 PM

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

**Jskid** · February 27th 2011, 02:05 PM

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

**Plato** · February 27th 2011, 02:50 PM

Have you considered
$\[ \left[ {\begin{array}{r} 1 \\ 1 \\ { - 1} \\ 0 \\ \end{array} } \right]~?$

**Jskid** · February 27th 2011, 04:42 PM

I didn't realize a row of 0s could be -r I thought it had to be r

**HallsofIvy** · February 28th 2011, 02:45 AM

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.

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