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I came across this in my matrix module, learning about number of solutions when the matrix has parameters. $$ \left[ \begin{array}{ccc|c} 1 & -1&1&1\\0&2&k^2&k^2-4\\0&0&k^2+k-2&k^2-4 \end{array} \right] $$

For there to be one solution, I need $k^2+k-2\ne0$, so that number of rows equal the number of unknowns. I came up with $(k+2)(k-1)\ne0$ therefore solutions $k\ne-2$ and $k\ne1$. But the worked solution to the question I'm doing has solutions $k\ne2$ and $k\ne-1$ (same values, but sign reversed).

Is there something I've missed when solving an inequation, or is there a mistake in the solution?

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  • $\begingroup$ Basically, to solve an inequation $f(x)\neq0$, you can solve the equation $f(x)=0$ and obtain a solution set $S$. These are the values you want to exclude, so the solutions of $f(x)\neq 0$ are the complement $S^c$ of $S$. This is actually what you did implicitely. $\endgroup$ – Taladris Aug 23 '14 at 2:14
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Factor $k^2-4$ as $(k+2)(k-2)$. You can see that, in conjunction with $(k+2)(k-1)$ that when $k=-2$, both expressions are zero, this results in a consistent system; the other option (as a result of factoring $k^2-4$, is for $k=2$.

And if I am not mistaken, you are correct about the second solution, $k=1$. I think there was a solution error for that.

Also, you are not finding ONE solution. This system, a 3X4, will have infinitely many, or none. What it is asking for are values of $k$ which make the system inconsistent. And, try to avoid doing math with the not equals symbol. It is generally a good way to write a solution, but it is often much more confusing and can lead you down the wrong path.

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  • $\begingroup$ Thanks for the answer. I have a couple of comments which are a bit linked... 1) The significant part of matrix (sorry I'm not sure how to call it in English), in order to find solution, doesn't include the fourth column. The fourth column is the result column. So it's true that if $k^2+k−2\ne0$ there will be one solution. 2) This is why I was talking about $k^2+k−2\ne0$, not $k^2-4$. Does this information change your answer? $\endgroup$ – antgel Aug 23 '14 at 11:45
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The fourth column is an augmented matrix, and no, that does not change my answer. Namely, because $k$ is included in the augmented column. If a value of $k$ makes that entire bottom row zero, including the augmented column entry, then there can be solutions between the first two rows. If such a value of $k$ made the bottom row zero, then it is consistent: $a*0+b*0+c*0=0$ for all $a, b, c$.

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  • $\begingroup$ I'll comment here as this is your latest answer. But regarding your answer above, I don't understand why you say there can't be one solution. If k2+k−2≠0 and there is no contradiction, then the number of unknowns (three) is equal to the number of nonzero rows (three), so surely that would imply one solution? Edited matrix mathjax to show it as augmented. $\endgroup$ – antgel Aug 24 '14 at 12:16
  • $\begingroup$ If your $3X4$ matrix represents a system of linear equations, then it can be thought of finding the intersections of three planes. $\endgroup$ – user170141 Aug 25 '14 at 0:43
  • $\begingroup$ Indeed. But why can't there be one solution? You said above "Also, you are not finding ONE solution. This system, a 3X4, will have infinitely many, or none." $\endgroup$ – antgel Aug 25 '14 at 19:32
  • $\begingroup$ The solutions to three real planes will be a line. Think about the intersection between two pieces of paper. They intersect at a line, not a point. $\endgroup$ – user170141 Aug 25 '14 at 20:24

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