simple systems of equations problem Choose h and k such that the system has 1) no solution, 2) a unique solution, and 3) many solutions. Give separate answers for each part.

x-3y=1,
2x+hy=k

For 1) and 3), isn't that impossible? And for 2) can't h and k be any real number?
You have that $x-3y=1$, $2x+hy=k$. When you multiply the first equation with $2$, you get $2x-6y=2$. Subtrating the two equations, you get, $2x+hy-(2x-6y)=k-2\Rightarrow (h+6)y=k-2$. If $h=-6$ and $k\neq 2$, you get $0y\neq 0$, so a contradiction. Therefore, for $h=-6, k\neq 2$, there are no solutions. If you take $h=-6$ and $k=2$, you get $0y=0$, so $y$ can have any value. If $h\neq -6$, it follows that $y=\frac{k-2}{h+6}$ and $x=1+3\frac{k-2}{h+6}$, therefore you get a unique solution. 

You could solve this, using determinants and matrix for systems, if you want a general algorithm.
You wish to analyze the system
\begin{array}{rcrcr}
x &amp; - &amp; 3\,y &amp; = &amp; 1 \\
2\,x &amp; + &amp; h\,y &amp; = &amp; k
\end{array}
To do so, note that $h\neq-6$ implies
$$
\DeclareMathOperator{rref}{rref}\rref
\begin{bmatrix}
1&amp;-3&amp;1\\
2&amp;h&amp;k
\end{bmatrix}
=
\begin{bmatrix}
1 &amp; 0 &amp; 3\,\frac{k-2}{h+6}+1 \\
0 &amp; 1 &amp; \frac{k-2}{h+6}
\end{bmatrix}
$$
This means that the system has a unique solution when $h\neq-6$. Do you see why?

Can you find the reduced row-echelon form of the system when $h=-6$?