# The full way of different result to a different presentation

1. Feb 20, 2009

### transgalactic

$$\begin{pmatrix} 0 & 2 & -3 & 2\\ 1 & a &-1 &-4 \\ 2&1 &-1 &-2 \\ 3 &0 &0 &-6 \end{pmatrix}$$
next
$$\begin{pmatrix} 1 &0 &0 &-2 \\ 0&2 &-3 &2 \\ 1& a &-1 &-4 \\ 2&1 &-1 &-2 \end{pmatrix}$$
next
$$\begin{pmatrix} 1& 0 &0 &-2 \\ 0&2 &-3 &2 \\ 0&a &-1 &-2 \\ 0&1 &-1 &2 \end{pmatrix}$$
next
$$\begin{pmatrix} 1 &0 &0 &-2 \\ 0&1 &-1 &2 \\ 0&0 &a-1 &-2-2a \\ 0&0 &-1 &-2 \end{pmatrix}$$
next

$$\begin{pmatrix} 1 &0 &0 &-2 \\ 0&1 &0 &4 \\ 0&0 &1 &2 \\ 0&0 &a-1 &-2-2a \end{pmatrix}$$
next
$$\begin{pmatrix} 1 &0 &0 &-2 \\ 0&1 &0 &4 \\ 0&0 &1 &2 \\ 0&0 &0 &-4a \end{pmatrix}$$
but when ill present the vectors as rows ill get
$$\begin{pmatrix} 0&1 &2 &3 \\ 2&a &1 &0 \\ -3&-1 &-1 &0 \\ 2& -4 &-2 &-6 \end{pmatrix}$$
next
$$\begin{pmatrix} 2&a &1 &0 \\ 0&1 &2 &3 \\ -3&-1 &-1 &0 \\ 2& -4 &-2 &-6 \end{pmatrix}$$
next
$$\begin{pmatrix} 2&a &1 &0 \\ 0&1 &2 &3 \\ 0&3a-2 &1 &0 \\ 0& -a-4 &-3 &-6 \end{pmatrix}$$
L_3 *(a+4) +L_4*(3a-2)=>L_4
$$\begin{pmatrix} 2&a &1 &0 \\ 0&1 &2 &3 \\ 0&3a-2 &1 &0 \\ 0& 0 &-8a+10 &-18a+12 \end{pmatrix}$$

see i get different results
so it does matter??

Last edited: Feb 20, 2009