I am struggling with solving systems of linear equations with unknown constants.
I can solve simple ones, such as:
In the following system, for which values of k would produce:
- Infinitely many solutions?
- No solution?
- A unique solution?
$$
x + ky = 1\\
kx + y = 1
$$
(where $k$ is some constant)
$$ \left[
\begin{array}{cc|c}
1&k&1\\
k&1&1
\end{array}
\right] $$
$$ \text{Operation: } R2 \leftarrow R2 – R1 $$
$$ = \left[
\begin{array}{cc|c}
1&k&1\\
0&1-k^2&1-k
\end{array}
\right] $$
Therefore:
a) If k = 1, there are infinitely many solutions:
$$ = \left[
\begin{array}{cc|c}
1&1&1\\
0&0&0
\end{array}
\right] $$
b) if k = -1, there are no solutions:
$$ = \left[
\begin{array}{cc|c}
1&-1&1\\
0&0&2
\end{array}
\right] $$
c) for any other values of k, we can do:
$$ \text{Operation: } R2 = \frac1{(1-k^2)}R2 $$
$$ = \left[
\begin{array}{cc|c}
1&k&1\\
0&1&\frac1{1+k}
\end{array}
\right] $$
But I can't figure out how to do this:
$$ = \left[
\begin{array}{ccc|c}
1&-2&3&2\\
1&1&1&k\\
2&-1&4&k^2
\end{array}
\right] $$
I try to put it into echelon form, but I end up with something that suggests there aren't any possible solutions:
$$ = \left[
\begin{array}{ccc|c}
1&-2&3&2\\
0&3&-2&k-2\\
0&0&0&k^2-k-6
\end{array}
\right] $$
Which was done by:
$$ R2 \leftarrow R2 – R1 $$
$$ R3 \leftarrow R3 – 2R1 $$
$$ R3 \leftarrow R3 – R2 $$
Best Answer
Assuming that you've row-reduced correctly:
We conclude that the system will be consistent if $k^2 - k - 6 = 0$ (so $k = 3,-2$) and therefore, for these $k$, the system have infinitely many solutions. In all other situations, the equation $$ 0 = k^2 - k - 6 $$ is a contradiction, so there are no solutions.
The system will never have a unique solution.