Cmunro said:
I shot off an email to my teacher, and he said that Gauss-Jordan is a much more complicated way to work it out. The hint is that there are an infinite number of solutions.
So I have canceled down the equations to look like this:
a=b
b=-c
b=-d
It seems very simple.. so I'm not sure. Apparently I should give the answer in terms of a, which would basically be
[a a]
[-a -a]
Using Gauss Jordan is a very simple thing, but it takes some time. And what is a, b etc? And the way I did it, the answers is in terms of t, a real parameter, but i took it out and put it in front of the matrix (multiplication with a scalar). And, yes, maybe Gauss jordan method is "overkill" in this problem, but when you have more complex equation systems to work on, this is the thing, nad using it with easier problems keep you "in shape"
Anyway this is a solution:
<br />
\left( \begin{array}{cc}1 & 0 \\ 1 & 2 \end{array}\right) \left( \begin{array}{cc}x_1 & x_2 \\ x_3 & x_4 \end{array}\right) = \left( \begin{array}{cc}x_1 & x_2 \\ x_3 & x_4 \end{array}\right) \left( \begin{array}{cc}2 & -1 \\ -1 & 2 \end{array}\right)<br />
<br />
\left( \begin{array}{cc} x_1 & x_2 \\ x_1 + 2x_3 & x_2 +2x_4 \end{array}\right) = <br />
\left( \begin{array}{cc} 2x_1 - x_2 & -x_1 + 2x_2 \\ 2x_3 - x_4 & -x_3 +2x_4 \end{array}\right)<br />
x_1 = 2x_1 - x_2
x_2 = -x_1 + 2x_2
x_1 + 2x_3 = 2x_3 - x_4
x_2 +2x_4 = -x_3 +2x_4
-x_1 + x_2 =0
x_1 - x_2 =0
x_1 + x_4 =0
x_2 + x_3 = 0
<br />
\left( \begin{array}{cccc|c} -1 & 1 & 0 & 0 & 0 \\ 1 & -1 & 0 & 0 & 0 \\ 1 & 0 &0 & 1 &0\\<br />
0&1&1&0&0 \end{array}\right)<br />
Adding and substractning multiples of the rows:
<br />
\left( \begin{array}{cccc|c} 1 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 1 &0 & 1 &0\\<br />
0&0&1&-1&0 \end{array}\right)<br />
now let x_4 = t , t \in \mathbb{R}
x_3 = x_4 \rightarrow x_3 = t , x_2 = -x_3 \rightarrow x_2 = - t
x_1 = x_2 \rightarrow x_1 = -t
<br />
X \equiv \left( \begin{array}{cc}x_1 & x_2 \\ x_3 & x_4 \end{array}\right) = t\left( \begin{array}{cc}-1 & -1 \\ 1 & 1 \end{array}\right)<br />
\end{equation}
