# Find values of a and b that make the system consistent

In the augmented matrix: $$\left(\begin{array}{rrr|r} 1 &-2 &4 & 7\\ 0 &a^2 - 1& a & 3\\ 0 &0 &b & -3 \end{array}\right).$$

How do I determine values for $a$ and $b$ that make the system consistent?

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We can't tell you what "it" is asking for unless you tell us exactly what question "it" asks. – Chris Eagle Oct 10 '11 at 20:34

"What values of $a$ and $b$ make the system consistent?" asks for all pairs of values for $a$ and $b$ that make the system consistent, not just a single one.

To find them, do Gaussian elimination, leaving expressions that involve $a$ and $b$ indicated. Be particularly careful when dividing by an expression involving $a$ or $b$, since you must ensure that the expression you divide by is nonzero.

For example, a first step, since the matrix is already in upper triangular form, might be to divide the last row by $b$ to make the $(3,3)$ entry into a $1$; but in order to "divide by $b$", you need $b\neq 0$. What happens in $b=0$ to the system? You need to consider that. Then, if you assume $b\neq 0$, then you can divide by $b$ and proceed from there.

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Makes sense - so is the idea to transform the matrix so that the rows contain leading ones rather than the unknowns? And since I cannot divide by zero, I can rule out a = 1 and a = -1? Is that it? – Jon Martin Oct 10 '11 at 20:44
@JonMartin: The idea is to transform it into a matrix in which you can just tell whether it is consistent or not consistent. You cannot divide by $a^2-1$ if $a=1$ or $a=-1$, but that does not mean the system is necessarily inconsistent under that circumstance. We know $b=0$ leads to an inconsistent system because if $b=0$, then the last row of the matrix reads $0\ 0\ 0 \mid\ -3$, which shows the system if inconsistent. But if $a=1$, then the second line reads $0\ 0\ 1 \mid\ 3$, and that, by itself, does not seem to pose a problem. So do Gaussian elimination (with care!) and see. – Arturo Magidin Oct 10 '11 at 20:52
Alright. One last thing - knowing that a can be 1 or -1, doesn't that mean b must be equal to -a under that circumstance for a consistent (but undeterminable) system? How would I write that in my solution? – Jon Martin Oct 10 '11 at 21:10
@JonMartin: You are being asked to write down all values that makes the system consistent. So first you would say somethign like "$b$ must be nonzero". Then "If $a=1$, then we must have ...", "If $a=-1$, then we must have ..."; "if $a\neq 1$ and $a\neq -1$, then...". That is, go through the cases and explain under those circumstances, what conditions you need to ensure the system is consistent. – Arturo Magidin Oct 10 '11 at 21:13
@JonMartin: If you are not quite clear yet, then I suggest the following: Try to work it out. Then post it as an answer! It's okay to post answers to your own question. That way, people can comment and let you know whether it is correct, or if there is a better way to say something (or if you said something you did not mean to say, or didn't say something that needs to be said). – Arturo Magidin Oct 10 '11 at 21:19

I think I have come up with an answer, I would appreciate it if it could be verified.

I observe that b must be non-zero for a consistent system. a can be any real number. If a = 1, then b must equal -1. If a = -1, then b must equal 1. There is a unique solution given these conditions where a != 1 and a != -1. There are infinitely many solutions where a = 1 and b = -1, or when a = -1 and b = 1.

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So, in summary: the system is consistent if: $a\neq \pm 1$ and $b\neq 0$ (in which case there is a unique solution); and if $a=\pm 1$ and $b=\mp 1$ (in which case there are infinitely many solutions); and in no other case. Yes, that seems right. – Arturo Magidin Oct 10 '11 at 21:45

I want to make sure I have the hang of this with one more example.

Given the matrix:

1 0 2 5 | 2

0 c c 0 | 1

0 0 c 0 | c

0 0 0 cd | c + d

I observe that c must be non-zero and that d must be non-zero. Given these conditions there is always a unique solution. Is this correct?

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Looks right to me. – Henning Makholm Oct 10 '11 at 22:42