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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}$,

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  • $\begingroup$ I think that if you are trying to solve a system of linear equations, using augmented matrix would be better. (I.e., you should add right-hand sides, too). $\endgroup$ – Martin Sleziak Aug 25 '15 at 10:38
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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.

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Hint: 1. No solution when: $$\frac { 1 }{ { k }^{ 2 } } =\frac { 2 }{ -2k } \neq \frac { 1 }{ k+2 } $$ it holds when $k=0$

  1. An infinite number of solutions when :$$ \frac { 1 }{ { k }^{ 2 } } =\frac { 2 }{ -2k } =\frac { 1 }{ k+2 } $$ it holds when $k=-1$
  2. Exactly one solution when:$$ \frac { 1 }{ { k }^{ 2 } } \neq \frac { 2 }{ -2k } $$ and it hold when $k\neq -1$
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  • $\begingroup$ can i get to kow if my row reduction form is correct? $\endgroup$ – shidiv Aug 24 '15 at 21:34
  • $\begingroup$ What you have done is not clear for me,what did you get?i gave hint,recomment to use from it $\endgroup$ – haqnatural Aug 24 '15 at 21:39
  • $\begingroup$ so you have given me a final answer i guess? $\endgroup$ – shidiv Aug 24 '15 at 21:55
  • $\begingroup$ it is a hint actually,but hope you can find from it all answers) $\endgroup$ – haqnatural Aug 24 '15 at 21:58
  • $\begingroup$ 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}$, $\endgroup$ – shidiv Aug 24 '15 at 22:38
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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 $0$,

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  • $\begingroup$ yes it is a matrix is my reduction form correct as mentioned in the previous comment? $\endgroup$ – shidiv Aug 24 '15 at 22:57
  • $\begingroup$ Please start using LaTeX, see this guide: meta.math.stackexchange.com/questions/5020/… Almost all of your posts (seven out of 8 answers you posted yesterday!) need heavy editing. $\endgroup$ – Hirshy Aug 25 '15 at 7:10

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