Find $\alpha$ such that $b$ is in the column space of $A$ 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.
