$$A = \begin{bmatrix}1&1&1\\3&2&1\\2&1&2\end{bmatrix}$$
$$s = \begin{bmatrix}15\\28\\23\end{bmatrix}$$
How to convert the above system into echelon form? Replace $A$ and $s$ with the correct values below:
$$\begin{bmatrix}1&A12′&A13′\\0&1&A23′\\0&0&1\end{bmatrix}$$
$$\begin{bmatrix}s1\\s2\\s3\end{bmatrix}$$
So first we write the system in the form of $$[A \qquad s]$$ as $$\begin{bmatrix} 1 & 1 & 1& 15 \\ 3 & 2 & 1& 28 \\ 2 & 1 & 2 & 23 \\ \end{bmatrix}$$ Now let's work on the last row to null the first element of it as $$r_3 \leftarrow r_3 - 2r_1$$ we get $$\begin{bmatrix} 1 & 1 & 1& 15 \\ 3 & 2 & 1& 28 \\ 0 & -1 & 0 & -7 \\ \end{bmatrix}$$ Now let's null the first element of the second row as $$r_2 \leftarrow r_2 - 3r_1 $$ $$\begin{bmatrix} 1 & 1 & 1& 15 \\ 0 & -1 & -2& -17 \\ 0 & -1 & 0 & -7 \\ \end{bmatrix}$$ Now let's set the diagonal elements to one starting from the last row $$r_3 \leftarrow \frac{1}{2}r_3$$ $$\begin{bmatrix} 1 & 1 & 1& 15 \\ 0 & -1 & -2& -17 \\ 0 & 0 & 1 & 5 \\ \end{bmatrix}$$ The second entry of the second row as $$r_2 \leftarrow -r_2$$ $$\begin{bmatrix}1 & 1 & 1& 15 \\ 0 & 1 & 2& 17 \\ 0 & 0 & 1 & 5 \\ \end{bmatrix}$$
Additional: If you want to solve it now, it is easy since we can easily embed the backward substitution method, i.e. $$s_3 = 5$$ $$s_2 + 2s_3= 17$$ so $$s_2 = 7$$ and $$s_1 + s_2 + s_3 = 15$$ so $$s_1 = 15 - (7+5) = 3$$