Linear Transforms & Matrices $T:R^4 -> R^3$ Linear Transform
This matrix is 
$[T]_{B2}^{B1}$ = A =\begin{pmatrix}1&2&3&4\\1&4&0&2\\2&2&9&10\end{pmatrix}
After elimination we get:
\begin{pmatrix}1&0&6&6\\0&1&-1.5&-1\\0&0&0&0\end{pmatrix}
Why the homogen solutions of this system is span{(6,-1,0,-1),(12,3,-2,0)}?
 A: Are you sure it's not
span{(6,-1,0,-1),(12,-3,-2,0)}
?
Well anyway, homogeneous solution means taking that matrix, multiplying it by an appropriate column vector and setting it equal to the zero vector of appropriate length.
Just take s=-1 and t = -2 below to get span{(6,-1,0,-1),(12,-3,-2,0)}.

If you want, you can say span{(-6,1,0,1),(-6,1.5,1,0)}. The point is that there are infinitely many solutions but each of those solutions is a linear combination of the two vectors (-6,1,0,1) and (-6,1.5,1,0). If I remember correctly the span of those vectors is ALL linear combinations of those two vectors.
That is, the solution set is not merely some of the linear combinations of those vectors but precisely all linear combinations of all those since t and s can be any real number.
Sometimes, t and s cannot be any real number and so we would have not the span of those two vectors but a subset of the span. I think.
For instance, maybe t represents time and so t is nonnegative. Then we would not have the span.
