<s>[INST] Determine the unique vector x in the row space for A, for which Ax = b Given the Vector $b = (9, -6, 3)^T$
$A =\begin{pmatrix}
         1 & 2 & 0 & 1\\
         -1 & -1 & -1 & 0 \\
         1 & 0 & 2 &-1
        \end{pmatrix}$
I know that the row space is $ \begin{pmatrix} 1 & 0 & 2& -1\end{pmatrix}$ , $ \begin{pmatrix} 0 & 1 & -1& 1\end{pmatrix}$
Not really sure how to proceed.
 [/INST] Let us take a vector $\ x=(x_1,x_2,x_3,x_4) $
For us to solve the given problem we need to solve the the system $\ Ax=b$
In our case$\ A=\pmatrix{1&2&0&1\\-1&-1&-1&0\\1&0&2&-1}$ and $\ b=\pmatrix{9\\-6\\3}$
$\ Ax=\pmatrix{1&2&0&1\\-1&-1&-1&0\\1&0&2&-1}*\pmatrix{x_1\\x_2\\x_3\\x_4}=\pmatrix{9\\-6\\3}$
After solving this system and taking $\ x_3,x_4$ as free variables you should get that$\ x=(3-2x_3+x_4,3+x_3-x_4,x_3,x_4)=(3,3,0,0)+x_3(-2,1,1,0)+x_4(1,-1.0,1)$
From this we conclude that our vector should be from vector space $\ (3,3,0,0)+span((-2,1,1,0),(1,-1.0,1))$
</s>