Let $A=\begin{bmatrix} 1 & 0 & -2 & 1\\ 0 & 4 & -1 & 0\\ 1 & 0 & -1 & 0\\ 2 & 4 & -4 & 1 \end{bmatrix} $ and let $b= \begin{bmatrix} 1\\ -3\\ 4\\ \alpha \end{bmatrix}.$ Find $\alpha$ such that $b$ is in the column space of $A$.
I’ve found $rref(A)= \begin{bmatrix} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & -1/4\\ 0 & 0 & 1 & -1\\ 0 & 0 & 0 & 0 \end{bmatrix}$ and thus Rank $A=3$.
Now, I know that the column space of $A$, $C(A)$, is the set of all linear combinations of the columns of $A$. And I know rank $A$. How can I relate all this information to find $\alpha$?
You have to find the RREF of the augemnted matrix
$A=\begin{bmatrix} 1 & 0 & -2 & 1&1\\ 0 & 4 & -1 & 0&-3\\ 1 & 0 & -1 & 0&4\\ 2 & 4 & -4 & 1&\alpha \end{bmatrix} $
And then see that for what value of $\alpha$ do you get rank of the augmented matrix to be $3$. (That is the last row is all zeroes). That would imply that you will have a solution.