I'm trying to prove this property of inverses in Artin's algebra text.
An element $a$ may have a left inverse or a right inverse, though it is not invertible.
I don't know if there is a general proof of this fact. What I'm ultimately trying to establish is that there exists a $b$ such that $ab = e$ but $ba \neq e$. Does it suffice to give an example of such an element?
For example, since $a$ can be any element, I can write a $1 \times 2$ matrix, \begin{align*} a = \begin{bmatrix} 2 & 3 \end{bmatrix}. \end{align*} Then, I have a right inversre, \begin{align*} b = \begin{bmatrix} x \\ y \end{bmatrix}, \end{align*} where $2x + 3y = 1$. Picking $y = 0$ and $x = \frac{1}{2}$ easily gives a right inverse, $\begin{bmatrix} \frac{1}{2} \\ 0 \end{bmatrix}$, so $ab = 1$, though it is not the case that $ba = 1$, since $ba$ isn't even defined. Similarly, $a$ is a left inverse of $b$, but not a right inverse, so neither $a$ nor $b$ are invertible.
Does this suffice as a proof, or should I aim for something more general?
My guess is that the context of that sentence is the study of rings. In that case, your example will not work, since you did not say which ring contains those two matrices (and I don't imagine which ring it could be).
You can take the ring of the linear maps from the space $L$ of sequences of real numbers into itself and define $f\colon L\longrightarrow L$ as$$f(a_1,a_2,a_3,a_4,\ldots)=(0,a_1,a_2,a_3,\ldots).$$It has a left inverse:$$\begin{array}{rccc}g\colon&L&\longrightarrow&L\\&(a_1,a_2,a_3,\ldots)&\mapsto&(a_2,a_3,a_4,\ldots),\end{array}$$but no right inverse.