Solving linear system of equations to obtain different classes of solution.

Question

Correct me if I am wrong. Find the value(s) of the constant $k$ such that the system of linear equations

$$\left\{\begin{array}{l} x + 2y = 1\\[2ex] k^2x − 2ky = k + 2 \end{array} \right.$$

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 $ \begin{bmatrix} 1 & 2 \\ k^2 & -2k \\ \end{bmatrix}$,

to $ \begin{bmatrix} 1 & 2 \\ 0 & -2k+2k^2 \\ \end{bmatrix}$,

Answer 1

Hint: 1. No solution when: $$\frac { 1 }{ { k }^{ 2 } } =\frac { 2 }{ -2k } \neq \frac { 1 }{ k+2 } $$ 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:$$ \frac { 1 }{ { k }^{ 2 } } \neq \frac { 2 }{ -2k } $$ and it hold when $k\neq -1$

Answer 2

Is your "row reduction form" supposed to be a matrix?

Here is how I would do this: From $x+ 2y= 1, x= 1- 2y$. Putting that into the second to get $k^2- 2^2y- 2ky= k+ 2$. From that, $2k(k+ 1)y= k^2- k- 2$ so $y= (k^2- k- 2)/(2k(k+ 1))$.

As long as the denominator is not $0$, that is, as long as k is not $0$ or $-1$, there is one solution. If the denominator is $0$, then there are either no solution (if the numerator is not also $0$) or an infinite number of solutions (if the numerator is also $0$). If $k= 0$, the numerator is $-2$ and if $k= -1$, the numerator is $0$,

Answer 3

You row reduction is wrong. We get $$\begin{array}{cc|c} 1&2&2\\ 0&-2k^2-2k&-k^2+k+2 \end{array}$$ which is equivalent to $$\begin{array}{cc|c} 1&2&2\\ 0&2k(k+1)&(k+1)(k-2) \end{array}$$

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.