How to determine if vector b is in the span of matrix A? Given a matrix A = \begin{bmatrix}
1 &2  &3 \\ 
4 &5  &6 \\ 
7 &8  &9 
\end{bmatrix}
Determine if vector $b$ is in $span(A)$
where 
$$
b = \begin{bmatrix}
1  \\ 
2  \\ 
4  
\end{bmatrix}
$$
 A: First we address $\mathrm{Span}(A)$,
$$\mathrm{Span}(A) = \left\{ \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ \end{array}\right] : \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ \end{array}\right] = a \left[\begin{array}{c} 1\\ 4\\ 7\\ \end{array}\right] + b \left[\begin{array}{c} 2\\ 5\\ 8\\ \end{array}\right] + c \left[\begin{array}{c} 3\\ 6\\ 9\\ \end{array}\right]. a,b,c \in \mathbb{R} \right\}.$$
From this definition we can see that asking if vector $\vec{b} \in \mathrm{Span}(A)$ is equivalent to asking if there exists a vector $\vec{x}$ such that $A\vec{x} = \vec{b}$,
because $$\vec{x} = \left[\begin{array}{c} x_1\\ x_2\\ x_3\\ \end{array}\right],$$
then $$A\vec{x} = \left[\begin{array}{ccc} 1 & 2 & 3\\ 4& 5 & 6\\ 7 & 8 & 9\\ \end{array}\right] \left[ \begin{array}{c} x_1\\ x_2\\ x_3 \end{array}\right] = x_1 \left[ \begin{array}{c} 1\\ 4\\ 7 \end{array}\right] + x_2 \left[ \begin{array}{c} 2\\ 5\\ 8 \end{array}\right] + x_3 \left[ \begin{array}{c} 3\\ 6\\ 9 \end{array}\right].$$
So if there exists $x_1, x_2, x_3$ such that the final line equals $\vec{b}$, then we know $\vec{b}$ is in $\mathrm{Span}(A)$. This means it suffices to ask if there exists a $\vec{x}$ such that $A\vec{x} = \vec{b}$.
This final equation is equivalent to the matrix equation:
$$\left[\begin{array}{ccc} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{array} \right] \vec{x} = \left[\begin{array}{c} 1\\ 2\\ 4\\ \end{array}\right],$$
which we can convert to the system:
$$\left[\begin{array}{ccc|c}
  1 & 2 & 3 & 1\\
  4 & 5 & 6 & 2\\
  7 & 8 & 9 & 4\\
\end{array}\right].$$
As you might have learned, we solve this system by row reduction (I used technology for this step, your instructor may require row reduction by hand):
$$\mathrm{RREF}\left(\left[\begin{array}{ccc|c}
  1 & 2 & 3 & 1\\
  4 & 5 & 6 & 2\\
  7 & 8 & 9 & 4\\
\end{array}\right]\right) = \left[\begin{array}{ccc|c}
  1 & 0 & 0 & \frac{-4}{3}\\
  0 & 1 & 0 & \frac{8}{3} \\
  0 & 0 & 1 & -1\\
\end{array}\right]. $$
This implies the vector $\vec{x}$ we seek is given by:
$$\vec{x} = \left[\begin{array}{c} \frac{-4}{3}\\ \frac{8}{3} \\ -1\\ \end{array} \right].$$
The existence of this $\vec{x}$ alone guarantees $\vec{b} \in \mathrm{Span}(A)$, but lets check our answer by computing $A\vec{x}$.
$$A\vec{x} = \left[\begin{array}{ccc} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{array} \right] \left[\begin{array}{c} \frac{-4}{3}\\ \frac{8}{3} \\ -1\\ \end{array} \right] = \left[\begin{array}{c} 1\\ 2 \\ 4\\ \end{array} \right], $$
as desired.
Some things to think about: are there any vectors in $\mathbb{R}^3$ that are not in Span($A$)? If so can you find them, if not can you justify it? Hope this answer helps!
A: Before doing any mathematical problems, one should know what each term in the problem means. (And so that you would be able to ask a better question.)
"$b$ is in $span(A)$" means the vector $b$ is a linear combination of the column (row) vectors of $A$.
A: You put the vector to the right (4th column) of the matrix and you do column reductions with respect to the first 3 columns. If the 4th column end up being zero it is in the span (and you may find which by doing the inverse of the column reductions).
Note that the property of the 4th column being in the span of the first 3 is preserved under the column reduction process (because it is invertible).
