System of equations $$\begin{align*} x -2y +3z &= 2\\ 2x -3y +8z &= 7\\ 3x -4y +13z &= 8 \end{align*}$$
In a augmented matrix, $3\times 4$ $$\left(\begin{array}{crc|c} 1 &-2 &3 &2\\ 2 &-3 &8& 7 \\ 3 &-4 &13& 8\\ \end{array}\right).$$
Above are the equations/rows. Using these row operations: a constant number times a row, interchanging/switching equations/row, and lastly adding two equations/rows together after multiplying one of them by a constant -- I am still unable to solve and end up having to cancel out a part of the diagonal of the matrix when trying to achieve a result similar or equal to the identity matrix. By cancel out I mean turn into 0 which makes it not recoverable using these operations. This is a problem when trying to reach a upper triangular matrix (row echelon or reduced row echelon but in this case whichever).
A similar situation seems to be the case with two unknowns and the same three number of equations
$$\begin{align*} x +y &= 1\\ 9x -y &= 0\\ x -19y &= -17 \end{align*}$$
The system is inconsistent: the second equation minus twice the first equation gives $y+2z=3$, but the third equation minus three times the second equation gives $2y+4z=2$. Those two equations are not consistent, so the system has no solutions. You can see that when doing Gaussian elimination: $$\begin{align*} \left(\begin{array}{crc|c} 1 &-2 &3 &2\\ 2 &-3 &8& 7 \\ 3 &-4 &13& 8\\ \end{array}\right) &\stackrel{R_2-2R_1}{\longrightarrow} \left(\begin{array}{rrr|r} 1 & -2 & 3 & 2\\ 0 & 1 & 2 & 3\\ 3 & -4 & 13 & 8 \end{array}\right) \stackrel{R_3-3R_1}{\longrightarrow} \left(\begin{array}{rrr|r} 1 & -2 & 3 & 2\\ 0 & 1 & 2 & 3\\ 0 & 2 & 4 & 2 \end{array}\right)\\ &\stackrel{R_3-2R_2}{\longrightarrow} \left(\begin{array}{rrr|r} 1 & -2 & 3 & 2\\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & -4 \end{array}\right) \stackrel{-\frac{1}{4}R_3}{\longrightarrow} \left(\begin{array}{rrr|r} 1 & -2 & 3 & 2\\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 1 \end{array}\right). \end{align*}$$ Both the coefficient and the augmented matrices are now in row echelon form: the first nonzero entry of each row is a $1$, and the $1$ in each row appears to the right of the $1$ in the previous row, and all rows of $0$s are at the bottom.
Since the rank of the reduced coefficient matrix is $2$, but the rank of the augmented matrix is $3$, the system is inconsistent. You can continue from here to get the reduced row echelon form, but it doesn't suddenly "fix" the problem: $$\begin{align*} \left(\begin{array}{crc|c} 1 & -2 & 3 & 2\\ 0 & 1 & 2 & 3\\ 0 & 0 & 0& 1 \end{array}\right)&\stackrel{R_1-2R_3}{\longrightarrow} \left(\begin{array}{crc|c} 1 & -2 & 3 & 0\\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 1 \end{array}\right) \stackrel{R_2-3R_3}{\longrightarrow} \left(\begin{array}{crc|c} 1 & -2 & 3 & 0\\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)\\ &\stackrel{R_1+2R_2}{\longrightarrow} \left(\begin{array}{ccc|c} 1 & 0 & 7 & 0\\ 0 & 1 & 2 & 0\\ 0 & 0 & 0 & 1 \end{array}\right). \end{align*}$$ This is in reduced row-echelon form (the added condition is that the leading $1$s in each row are the only nonzero entries in their columns), but again you see that it corresponds to an inconsistent system.
Now, suppose that the last equation was $3x-4y+13z=12$, instead of $8$. Then in the third step of the reduction above we would have gotten the matrix $$\left(\begin{array}{crc|c} 1 & -2 & 3 & 2\\ 0 & 1 & 2 & 3\\ 0 & 2 & 4 & 6 \end{array}\right)\stackrel{R_3-2R_2}{\longrightarrow} \left(\begin{array}{crc|c} 1 & -2 & 3 & 2\\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 0 \end{array}\right).$$ This is a perfectly fine row-echelon form matrix; it's just that you will end up with one free variable and infinitely many solutions. You can then proceed to do the reduced row echelon form by adding twice the second row to the first row and you would get $$\left(\begin{array}{ccc|c} 1 & 0 & 7 & 8\\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 0 \end{array}\right)$$ which gives the solutions $$\begin{align*} x &= 8 - 7t\\ y &= 3 - 2t\\ z &= t \end{align*},\qquad t\text{ arbitrary.}$$