Choose h and k such that the system has a solution, a unique solution and many solutions. Im learning linear algebra, and im tasked with choosing $h$ and $k$ such that this system:
$$
\begin{cases}
x_1+hx_2=2\\
4x_1+8x_2=k\\
\end{cases}
$$
Has (a) no solution, (b) a unique solution, and (c) many solutions.
First i made an augmented matrix, then performed row reduction:
$$ \left[
      \begin{array}{cc|c}
        1&h&2\\
        4&8&k
      \end{array}
    \right] \sim  \left[
      \begin{array}{cc|c}
        1&h&2\\
        0&8-4h&k-8
      \end{array}
    \right]$$
Continuing row reduction, i get:
$$\sim \left[
      \begin{array}{cc|c}
        1&0&\frac{k-8}{2(h-2)}+\frac{k}{4}\\
        0&1&\frac{k-8}{8-4h}
      \end{array}
    \right]$$
Im not sure how to go about solving the problem with the matrix i end up with? Or if im going about the problem in the correct manner?
 A: It is better not to do row reduction when you have unknown parameters such as $h$ and $k$ in the matrix...for example suppose $h=2$, what happens to your last matrix?
Since you have a very simple system you can derive the answers directly, by considering the following:


*

*The system will have no solution when the coefficient matrix rows are linearly dependent (one row is a multiple of the other), BUT the augmented matrix rows are linearly independent.

*The system will have exactly one solution when the coefficient matrix rows are linearly independent.

*The system will have many solutions when the augmented systems rows are linearly dependent. 

A: It may make it easier to see what happens if you leave the ratio in the first row of the "constants column" in your reduced matrix over a common denominator:
$$ \left[\begin{array}{cc|c}1&0&\frac{16 - hk}{4(2-h)}\\0&1&\frac{k-8}{4(2-h)}\end{array}\right]$$
As pointed out by Christiaan Hattingh, there's trouble if $ \ h \ = \ 2 \ $ for any value of $ \ k \ $ .  So the system of equations is consistent (one solution) if $ \ h \ \neq \ 2 \ $ .  If $ \ h \ = \ 2 \ $ and $ \ k \ = \ 8 \ $ , we would have $ \ \frac{0}{0} \ $ in both entries of the constant column.  This would be interpreted to mean that both variables take on "indeterminate" values.  The original matrix looks like:
$$ \left[
      \begin{array}{cc|c}
        1&2&2\\
        4&8&8
      \end{array}
    \right] \ \ , $$
in which the second row is a multiple of the first; thus we have a "dependent" system with the single line equation $ \ x + 2y \ = \ 2 \ $ .  The remaining case is the one for which  $ \ h \ = \ 2 \ $ and $ \ k \ \neq \ 8 \ , $ giving us a matrix
$$ \left[
      \begin{array}{cc|c}
        1&2&2\\
        4&8&\neq8
      \end{array}
    \right] \ \ , $$
which is now an "inconsistent" system, one with no solutions. (In the row-reduced augmented matrix at the start of this post, the entries in the constants column become two ratios having a non-zero number divided by zero.)
A: Write the system as $A_h x = b_k$.
If $\det A_h \neq 0$ then there is a unique solution $x = A_h^{-1} x_k$.
If $\det A_h = 0$, then there no solution if $b_k \notin {\cal R} A_h$, and many solutions otherwise (since $\ker A_h \neq \{0\}$).
Since $\det A_h = 8-4h$, we see that $\det A_h \neq 0$ iff $h \neq 2$.
If $h=2$, then ${\cal R} A_h = \operatorname{sp}\{ \binom{2}{8} \}$, hence 
$b_k \notin {\cal R} A_h$ iff $k\neq 8$
