Question

In problem set 6.2, why isn't
[x, y, z]= x[12, 0, 0] + y[3, 1, 0] +z[1, 0, 1]. Is it because the x coefficient is the pivot so y and z are the free variables?

Answer 1

Are you talking about these problem sets?
http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/assignments/

I can't find the question you are talking about, please link me to the page where the pdf is.

Answer 2

Okay, I found the problem set! http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/column-space-and-nullspace/MIT18_06SCF11_Ses1.6prob.pdf

The reason the solution is \[\left[\begin{matrix}12\\ 0\\0\end{matrix}\right] + x\left[\begin{matrix}3\\1\\0\end{matrix}\right] + y\left[\begin{matrix}1\\0\\1\end{matrix}\right]\] is because that x and y are not the same x and y from the equation x - 3y - z = 12. So, a perfectly good answer to the problem would have been\[\left[\begin{matrix}12\\ 0\\0\end{matrix}\right] + c\left[\begin{matrix}3\\1\\0\end{matrix}\right] + d\left[\begin{matrix}1\\0\\1\end{matrix}\right]\]

While it is a bit confusing now, I think Strang clarifies everything over the next couple of lectures. The \[\left[\begin{matrix}12\\ 0\\0\end{matrix}\right]\] part of the answer is later called the particular answer or \[x_{particular}\] and the other two parts:\[\ x\left[\begin{matrix}3\\1\\0\end{matrix}\right] + y\left[\begin{matrix}1\\0\\1\end{matrix}\right]\] are called special solutions. Notice that they are each in the nullspace of the original matrix...

\[x-3y-z \Rightarrow \left[\begin{matrix}1&-3&-1\end{matrix}\right] \]

Multiplying \[\left[\begin{matrix}1&-3&-1\end{matrix}\right] \left[\begin{matrix}3\\1\\0\end{matrix}\right] = \left[\begin{matrix}0\end{matrix}\right]\]You can multiply the matrix by any combination of the special solutions, and you will still land in the nullspace (matrix of all zeroes).