Using a matrix to answer Ax=b question Use Matrix $A$ to answer the following.
This is matrix  $A$
Let $A= \begin{bmatrix}
1 & 3 & 0&  3 \\
-1&-1& -1&   1 \\
0&  -4&   2&   -8 \\
2 &0 & 3& -1
\end{bmatrix}
$
Turning into echelon form I get
 $A= \begin{bmatrix}
1 & 3 & 0&  3 \\
0& 2& -1&  4 \\
0&  0&   0&   5 \\
0& 0 & 0 & 0
\end{bmatrix}
$
I am having a tough time answering these question
1. How many rows of $A$ contain a pivot? My answer $3$
2. Does the equation $Ax=b$ have a solution for each $b$ in $\mathbb{R}^4$?
My answer: no because my last row is all zero unless the last entry in $\mathbb{R}^4$ is zero.
3. Can each vector be written as a linear combination of the columns of the matrix $A$ above?
4. Do the columns of $A$ span $\mathbb{R}^4$?
I am not sure about the last two.
 A: I won't solve the problems for you, because you already have the solutions to the last two problems: they are merely rewording of the previous parts. Instead I will try to help you understand the meaning of these two concepts better: span, and linear combinations.


*

*Linear combination: any vector of the form $c_1+\dots+c_kv_k$ is called a linear combination of $v_1, \dots, v_k$. Here $c_i$ are numbers.

*Span: Span of a set of vectors $v_1, \dots, v_k$ is the set of all of their linear combinations; that is to say, $\mathrm{Span}\{v_1, \dots, v_k\}=\{c_1v_1+\dots+c_kv_k\ \text{ for all real (or complex) numbers } c_i\}$.


Now let's look at part $4$ for instance. Columns of $A$ are $4$ vectors, say $A_1, \dots, A_4$, in $\mathbb{R}^4$. The question is asking whether 
$$\mathrm{Span}\{A_1, \dots, A_4\}=\mathbb{R}^4.$$
That is to say, whether any $b$ in $\mathbb{R}^4$ can be written as 
$$x_1A_1+\dots+x_4A_4=b.$$
But this is equivalent to asking whether the system $Ax=b$ is solvable for all choices of $b$, a question that you have already answered.
