[Math] Choose h and k such that the system has a solution, a unique solution and many solutions.

Question

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?

Answer 1

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:

1. 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.
2. The system will have exactly one solution when the coefficient matrix rows are linearly independent.
3. The system will have many solutions when the augmented systems rows are linearly dependent.