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I have this system

$ \left[ \begin{array}{ccc|c} -3&0&-1&0\\ 2&0&0&0\\ 2&0&0&0\\ \end{array} \right] $ Is there something you can tell from looking at this straight away?

When I reduce it I get this:

$ \left[ \begin{array}{ccc|c} 1&0&-1&0\\ 0&0&1&0\\ 0&0&0&0\\ \end{array} \right] $

I thought I read that if you get a statement that does not make sense, then the system is inconsistent, so $ 1 = 0$ does not make sense to me. Does that not mean it's inconsistent?
However, I thought that $x_3$ could equal $0$, and so could $x_1$ and basically, $x_2$ could equal anything. Still not sure, i reduced it further to:
$ \left[ \begin{array}{ccc|c} 1&0&0&0\\ 0&0&1&0\\ 0&0&0&0\\ \end{array} \right] $

From this, am I correct now in thinking that:
$ x_1 = r\\ x_2 = s \\ x_3 = t $

I am not sure of why I'm thinking like this, in terms of free variables etc. If I'm right, I wouldn't mind some reasoning as to why this is so, or being correct if I am wrong. Thanks.

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  • $\begingroup$ How did you do the final step of reduction? $\endgroup$ – platty Aug 2 '17 at 23:17
  • $\begingroup$ Opps, I see where you are going with this. The 1 should remain in the second row. Sorry about that. Let me think for a minute. $\endgroup$ – Bucephalus Aug 2 '17 at 23:20
  • $\begingroup$ Now that I have corrected my last reduction, I'm still not really sure what this means. it looks like $x_2$ can equal anything, Therefore, $x_1$ and $x_3$ can equal anything? Is that correct? $\endgroup$ – Bucephalus Aug 2 '17 at 23:24
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Remember that the entries on the left are coefficients of variables. So from your final reduction, we would end up with the RREF: $$\left[\begin{array}{ccc|c} 1&0&0&0\\ 0&0&1&0\\ 0&0&0&0\\ \end{array}\right]$$

The first line tells us that $1x + 0y + 0z = 0$, so $x = 0$. The second line tells us that $0x + 0y + 1z = 0$, i.e. $z=0$. But we have $y$ as a free variable, so our solution is $(0,y,0)$.

Note that we can tell that $y$ is free by observing the original matrix - there is no coefficient in its column, so its value doesn't affect our final solution.

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  • $\begingroup$ Oh, yes I see now. Thanks for your explanation @S.Ong. $\endgroup$ – Bucephalus Aug 2 '17 at 23:27
  • $\begingroup$ Say it did just turn out to be like I originally had it, with a single 1 in the first column of the first row. That would then mean that $x_1=0, x_2=r$ and $x_3=s$. Would that be right? $\endgroup$ – Bucephalus Aug 2 '17 at 23:29
  • $\begingroup$ Yep, that's it! :) $\endgroup$ – platty Aug 2 '17 at 23:32
  • $\begingroup$ Ok, thankyou @S.Ong. $\endgroup$ – Bucephalus Aug 2 '17 at 23:32
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$\left[ \begin{array}{ccc|c} -3&0&-1&0\\ 2&0&0&0\\ 2&0&0&0\\ \end{array} \right]$

What do you know straight away?

$(0,0,0)$ is in the solution set. The right column being all $0$ tells you this.

With the second row and the 3rd row identical or the right part of the matrix, this matrix is singular.

In fact the rank is 2, the nullity is 1, and there is a "line" of solutions.

The second column all zero's tells you the very same thing. Furthermore, $x_2$ can be anything.

And the structure of rows 2 and 3 tells you that $x_1$ must equal $0.$

this is more than enough information to let you know the solution set is $(0,t,0)$

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  • $\begingroup$ Aaah, very insightful. This is the kind of thing I was looking for also. Do you mean by "line" that the solutions form a line in 3-space? @DougM $\endgroup$ – Bucephalus Aug 3 '17 at 0:30
  • $\begingroup$ yes, with the rank of the kernel equal to 1, (and the information that at least one solution exists), then all of the solutions will lie on a line in $\mathbb R^3.$ If the nullity equaled 2, then we might say that there is a plane of solutions. $\endgroup$ – Doug M Aug 3 '17 at 0:36
  • $\begingroup$ Thanks @DougM. I will keep an eye out for you next time. $\endgroup$ – Bucephalus Aug 3 '17 at 0:43

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