Q:

If $A$ is a matrix $10$ by $12$ and $A x=b$ is solvable for every $b$, then th column space of $A$ is?

If $A$ is a matrix $10$ by $12$ and $A x=b$ is solvable for every $b$, then th column space of $A$ is?
The column space of $A$ is the whole of $R^m$ which is $12$ is this answer the right answer ?

A:

According to the hypothesis of your problem, for all fixed vector
$
b=\begin{pmatrix} b_1 \\\vdots \\ b_j\\ \vdots \\b_{10} \end{pmatrix}
$
in $\mathbb{R}^{10}$ exists a vector
$
x=\begin{pmatrix} x_1 \\\vdots \\ x_j\\ \vdots \\x_{12} \end{pmatrix}
$
in $\mathbb{R}^{12}$ such that the equality below is true
$$
\begin{pmatrix}
A_{11} & \ldots & A_{1j} & \ldots & A_{1\,12}
\\
\vdots & \ddots & \vdots & \ddots & \vdots
\\
A_{i1} & \ldots & A_{ij} & \ldots & A_{i\,12}
\\
\vdots & \ddots & \vdots & \ddots & \vdots
\\
A_{10\,1} & \ldots & A_{10\,j} & \ldots & A_{10\,12}
\end{pmatrix}
\begin{pmatrix} x_1 \\\vdots \\ x_j\\ \vdots \\x_{12} \end{pmatrix}
=
\begin{pmatrix} b_1 \\\vdots \\ b_j\\ \vdots \\b_{10} \end{pmatrix}
$$
By performing the multiplication of the matrix $A$ by the column matrix $x$ we have that the following statement is valid: for all fixed $b\in\mathbb{R}^{10}$ there are numbers $x_1$, $\ldots$, $x_j$, $\ldots$, $x_{12}$ such that
$$ 
\begin{pmatrix}
x_1\cdot A_{11}+& \ldots &+x_j\cdot A_{1j}+ & \ldots &+x_{12}\cdot A_{1\,12}
\\
\vdots & \ddots & \vdots & \ddots & \vdots
\\
x_1\cdot A_{i1}+ & \ldots &+x_j\cdot A_{ij}+ & \ldots &+x_{12}\cdot A_{i\,12}
\\
\vdots & \ddots & \vdots & \ddots & \vdots
\\
x_1\cdot A_{10\,1}+ & \ldots & +x_j\cdot A_{10\,j} +& \ldots &+x_{12}\cdot A_{10\,12}
\end{pmatrix}
=
\begin{pmatrix} b_1 \\\vdots \\ b_j\\ \vdots \\b_{10} \end{pmatrix}
$$
Let the vector $A_j$ be the j-th column of matrix $A$.The last statement can then be rewritten as follows. For all vector $b\in\mathbb{R}^{10}$ there are real numbers $x_1$, $\ldots$, $x_j$, $\ldots$, $x_{12}$ such that
$$
x_1 \begin{pmatrix} A_{11} \\\vdots \\ A_{i1}\\ \vdots \\A_{10 \,1} \end{pmatrix}
+\ldots +
x_j \begin{pmatrix} A_{1j} \\\vdots \\ A_{ij}\\ \vdots \\A_{10 \,j} \end{pmatrix}
+\ldots +
x_{12} \begin{pmatrix} A_{1\,12} \\\vdots \\ A_{i\,12}\\ \vdots \\A_{10 \,12} \end{pmatrix} 
=
\begin{pmatrix} b_{1} \\\vdots \\ b_{j}\\ \vdots \\b_{10} \end{pmatrix}
$$
Since the vector $ b $ is arbitrary, it follows that the columns $A_1,\ldots, A_{12} $ of $ A $ are vectors that generate the space $\mathbb {R}^{10} $.

