For what values of k and h does this system of equations have a unique solution? Here's my system of equations:
$x−3y+2z=5$
$2x−5y−3z=9$
$−x−y+kz=h$
So I have $
\begin{bmatrix}
 1 &  -3 & 2 & 5 \\\\
 2 &  -5 & -3 & 9 \\\\
 -1 &  -1 & k& h 
\end{bmatrix}$
When I row reduce, I get: $
\begin{bmatrix}
 1 &  0 & -19 & 2\\\\
 0 &  1 & -7 & -1\\\\
 0 &  0 & k-26 & h+1
\end{bmatrix}$
1) has a unique solution.
2) has infinite number of solutions.
3) has no solution.
Not really sure where to go from here...
Any suggestions much appreciated!
Thanks, Alex
 A: Well let's think about this. First of all, when will there be no solution? There will be no solution when $k = 26$ and when $h$ is not equal to $-1$. Why is this? This is because we can't have $0x_{3}$ possibly equal any value greater than $0$ or less than $0$. This would mean that the system was inconsistent. There will be infinite solutions when $k = 26$ and $h = -1$. This is because anything multiplied by zero will in fact be 0. And there is an infinite amount of numbers to multiply by zero to get zero. Finally, there will be a unique solution when $k$ is not equal to $26$. In this case, the variable h can be anything.
A: Consider two matrix:
$$A=\begin{bmatrix}
 1 &  0 & -19 \\\\
 0 &  1 & -7 \\\\
 0 &  0 & k-26 
\end{bmatrix}$$
And $$B=\begin{bmatrix}
 1 &  0 & -19 & 2\\\\
 0 &  1 & -7 & -1\\\\
 0 &  0 & k-26 & h+1
\end{bmatrix}$$.
The linear system has solutions when $rk A=rk B$ (for the Rouché-Capelli's theorem) and the dimension of solution is given by: $\infty^{n-r}$ where $n$ is the number of unknowns and r is the $rk$ of the matrix. Note that when $k=26$ and $h\neq 1$ there aren't solutions. When $k=26$ and $h=1 1$ there are infinite solutions. When $k\neq 26$ there's a only solution 
