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

$\{\begin{array}{l}x+2y=1\\ {k}^{2}x-2ky=k+2\end{array}$

has:

1. No solution

2. An infinite number of solutions

3. Exactly one solution

Answer:

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 $\left[\begin{array}{cc}1& 2\\ 0& -2k+2{k}^{2}\end{array}\right]$

$\{\begin{array}{l}x+2y=1\\ {k}^{2}x-2ky=k+2\end{array}$

has:

1. No solution

2. An infinite number of solutions

3. Exactly one solution

Answer:

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

Beginner2022-06-22Added 18 answers

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

$\begin{array}{ccc}1& 2& 2\\ 0& 2k(k+1)& (k+1)(k-2)\end{array}$

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.

$\begin{array}{ccc}1& 2& 2\\ 0& -2{k}^{2}-2k& -{k}^{2}+k+2\end{array}$

which is equivalent to

$\begin{array}{ccc}1& 2& 2\\ 0& 2k(k+1)& (k+1)(k-2)\end{array}$

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

Beginner2022-06-23Added 8 answers

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 $k\ne -1$

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 $k\ne -1$