Help in linear algebra - system of equation with unknown variable I have this exercise and this type always confuses me.
I have the following system:

And it asks me to write the matrix, and solve it using Gauss Jordan. After that, I have to see for which values of h the system is solvable.
In the end, solve it for h in R
I solved it like this:

Now the only thing i can say in my opinion is that the system is not solvable for $h=0$ because in some steps we would be dividing by 0.
Also, if $ h=1 $ then we have an unique solution.
I know how to solve them, but i don't understand where to look for clues of what would be solvable and what not ?
Is it enough to look in the end at the result , or also keep track of the steps and be careful where the system might have infinite solutions , one solution, or impossible ?
Thank you
 A: When all the operations are justified, that is, when you haven't divided by $0$, there is a unique solution, and your calculation have produced it.  (I suppose, I haven't checked your work in full detail.)  So, you have to worry about the cases where you've divided by $0$.  
One case is $h=0$, but if you substitute that into the original system of equations, you see at once that $x=y=z=0$ is the only solution.  The other case is when you divide by $h^3+h+1$.  If that equals $0$ the system cannot possibly have a solution.  Look at the matrix where you first want to divide by $h^3+h+1.$
The third row would require $0x+0y+0z=h(h+1)$.  If $h^3+h+1=0$, then $h\neq0,-1$ so this equation can't hold.
A: The determinant of the matrix of coefficients is $$1(2h)+2h(h^2)+2=2(h^3+h+1)$$
Thus the system has a unique solution iff $$h^3+h+1\ne0$$
You have solved system correctly. 
A: The system is solvable for all $h\in\mathbb R-{h_0}$ where $h_0$ is a real root of $h^3+h+1=0$.

Added: As discussed in comments that for $h=0$ the system has trivial solution (can also be obtained from the last step reduced form of matrix).

For investigation at $h_0$ you just need to clear the fractions first from last step reduced form of matrix by multiplying each row with $h^3+h+1$. At $h=h_0$ $rank(A)=0$ and $rank(A:b)=3$ indicating no solution case.


