Consider the following two types of multiplication (Please, Read the whole question):
$$
\begin{bmatrix}
a & x & c \\
d & e & f \\
\end{bmatrix}. \begin{bmatrix}
g\\
h\\i
\end{bmatrix}
$$
and,
$$ \begin{bmatrix} a & x \\ d & e \\ f & g \end{bmatrix}. \begin{bmatrix} g\\ h \end{bmatrix} $$
Now, there's a rule in linear algebra that states:
"$T$ maps $\mathbb{R}^n$ onto $\mathbb{R}^m$ iff the columns of $A$ span $\mathbb{R}^m$"
Given the rule, the former matrix multiplication we see the fact that the multiplication results in the transformation of a three dimensional vector into two dimensions.
But, how is this possible, when the columns of the transformation matrix does not span $\mathbb{R}$ because the matrix is linearly dependent??
Similarly, the columns of the transformation matrix of the second multiplication is also not spanning $\mathbb{R}^3$ yet the vector of $\mathbb{R}^n$ with $n=2$ is being transformed to $\mathbb{R}^3$. How?
In the first case your matrix is $A\in \mathbb{R}^{2\times 3}$ and the induced homomorphism is $$f_A:\mathbb{R}^3\to \mathbb{R}^2, \quad f_A(x)=Ax$$
The columns of $A$ are never linearly independent, only $2$ can be since they are in $\mathbb{R}^2$. If two columns are, a $2$ dimensional subspace of $\mathbb{R}^3$ is being mapped onto $\mathbb{R}^2$ and one dimension is "lost information". The vectors "lost" are being mappend into $(0,0)^T$, which is called the kernel of $f_A$ or $A$.
In the second case $B\in \mathbb{R}^{3\times 2}$ and $$f_B: \mathbb{R}^2\to \mathbb{R}^3,\quad f_B(x)=Bx$$ Here if the columns are linearly independent then the entire space $\mathbb{R}^2$ is mapped onto a 2 dimensional subspace inside $\mathbb{R}^3$. The image is a plane and the rest isn't being "touched" by $f_B $.