Makayla Boyd

2022-06-21

Correct me if I am wrong. Find the value(s) of the constant k such that the system of linear equations
$\left\{\begin{array}{l}x+2y=1\\ {k}^{2}x-2ky=k+2\end{array}$
has:
1. No solution
2. An infinite number of solutions
3. Exactly one solution
so the first step is to get row reduction form, which is:
from $\left[\begin{array}{cc}1& 2\\ {k}^{2}& -2k\end{array}\right]$,
to $\left[\begin{array}{cc}1& 2\\ 0& -2k+2{k}^{2}\end{array}\right]$

Jayce Bates

You row reduction is wrong. We get
$\begin{array}{ccc}1& 2& 2\\ 0& -2{k}^{2}-2k& -{k}^{2}+k+2\end{array}$
which is equivalent to
$\begin{array}{ccc}1& 2& 2\\ 0& 2k\left(k+1\right)& \left(k+1\right)\left(k-2\right)\end{array}$
From here we see that there is no solution iff $k=0$, an infinite number of solutions iff $k=-1$ and else there is exactly one solution.

Feinsn

Hint:
1. No solution when:
$\frac{1}{{k}^{2}}=\frac{2}{-2k}\ne \frac{1}{k+2}$
it holds when $k=0$
2. An infinite number of solutions when:
$\frac{1}{{k}^{2}}=\frac{2}{-2k}=\frac{1}{k+2}$
it holds when $k=-1$
3. Exactly one solution when:
$\frac{1}{{k}^{2}}\ne \frac{2}{-2k}$
and it hold when $k\ne -1$

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