Find the number of values of $k$ for which the system of equations has no solution: $$(k+1)x+8y=4k$$ $$kx+(k+3)y=3k-1$$
This is the augmented matrix: $$\begin{bmatrix} k+1 & 8 & 4k \\ k & k+3 & 3k-1 \end{bmatrix}$$
$R_2 \rightarrow R_2-\frac{k}{k+1}R_1$
$$\begin{bmatrix} k+1 & 8 & 4k \\ 0 & k+3-\frac{8k}{k+1} & 3k-1-\frac{4k^2}{k+1} \end{bmatrix}$$
If in the last row, the coefficient of $y$ where to become zero, keeping the constant term non zero, then there will be no solution. That is, $$k+3-\frac{8k}{k+1}=0\text{ and } 3k-1-\frac{4k^2}{k+1}\neq 0$$
The only solution is $3$. But the answer says there are 3 solution for $k$.
Given two linear equations $ax+by=h,\ cx+dy=k,$ the only way there can fail to be a unique solution is for the determinant $ad-bc$ to be zero. For your system, the determinant works out as $(k-1)(k-3).$ So only $k=1,3$ could lead to nonexistent or nonunique solutions. In your system, $k=3$ is the matrix with rows [4,8,12],[3,6,8] and that has no solutions. On the other hand if $k-1$ the matrix has rows [2,8,4],[1,4,2] and one is the double of the other so the system is equivalent to the single line $x+4y=2$ so infinitely many solutions in this case.
One has to be careful when using gauss-Jordan elimination on matrices with variable entries. Any time one divides one has to separately consider any case where a denominator becomes zero. Maybe it is better to use "integer pivots", i.e. do not divide but instead use multiples of rows to get the zeros, and then look at things at the end. However for your problem, if you're willing to accept the determinant fact above, you can just plug in any values making a zero determinant and consider those matrices of constants one at a time.
After changing row 2 to $(k+1)R_2-kR_1$ the new row 2 is $$[0,(k-1)(k-3),-(k-1)^2].$$ Then after changing row 1 to $(k-1)(k-3)R_1-8R_2,$ the new row 1 is $$[(k-1)(k-3)(k+1),0,4(k-1)(k-2)(k+1)].$$ But to interpret this final matrix one still needs to consider the three numbers $-1,1,3$ which make one of the now diagonal entries zero. It will be found that at $k=-1$ one gets a unique solution $x=3,y=-1/2$ to the original system, but this manipulated system now misleads one into thinking there are infinitely many solutions in the $k=-1$ case, which is false.