# Systems of equations - consistency

1. Mar 1, 2005

### Benny

Q. Determine the conditions on a, b, c so that the following systems are consistent and find any solutions.

a)
$$\begin{array}{l} x + 2y - 3z = a \\ 3x - y + 2z = b \\ x - 5y + 8z = c \\ \end{array}$$

b)
$$\begin{array}{l} x + 2y + 4z = a \\ 2x + 3y - z = b \\ 3x + y + 2z = c \\ \end{array}$$

Firstly, the 'solutions' that I end up getting seem to either require me to introduce a parameter(for the first system) for one of the variables or the answer is in terms of a, b and c(for the seond systrem).

I am really unsure about how to do this question. For part a I wrote the corresponding augmented matrix and got down to what I think is it's reduced echelon form.

$$\left( {\left| {\begin{array}{*{20}c} 1 & 2 & { - 3} \\ 0 & { - 7} & {11} \\ 0 & 0 & 0 \\ \end{array}} \right|\begin{array}{*{20}c} a \\ {b - 3a} \\ {c + 2a - b} \\ \end{array}} \right)$$

So the condition need for the system to be consistent that c+2a-b = 0 right? I'm not sure what else needs to be said to determine the conditions for a, b and c so that the system is consistent. Also, when I solved the equation I let z = s where s is a real number so that I got:

$$\left( {x,y,z} \right) = \left( {\frac{{5a + 2c + s}}{7},\frac{{11s + a - c}}{7},s} \right)$$

For part b I went further so as to reduce the augmented matrix to row reduced echelon form:

$$\left( {\left| {\begin{array}{*{20}c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}} \right|\begin{array}{*{20}c} {\frac{{140b - 259a - 14c}}{{35}}} \\ {\frac{{133a - 80b + 9c}}{{35}}} \\ {\frac{{7a - 5b + c}}{{35}}} \\ \end{array}} \right)$$

Now the 'problem' with this is that none of the rows in the coefficient matrix are zero. So no matter what I try to do with the overall augmented matrix, it seems that the system is always consistent. Can this possibly be right? Also, from the augmented matrix the 'solution' to the system would be:

$$\left( {x,y,z} \right) = \left( {\frac{{140b - 259a - 14c}}{{35}},\frac{{133a - 80b + 9c}}{{35}},\frac{{7a - 5b + c}}{{35}}} \right)$$

Now that just looks way too complicated to be correct. Can someone please help me out with this question? Any help would be great thanks.

Last edited: Mar 1, 2005
2. Mar 1, 2005

### EvLer

For (a) it is sufficient to say that condition is c+2a-b = 0.
For (b) you have this: z = (7a - 5b + c)/35. And if you think about it, it shows that there a, b, c can be chosen arbitrarily, i.e. they can be any number, so a, b, c are not dependent on each other, unlike (a), where you can say that b = c + 2a and so forth, where each variable is dependent on other two.

Last edited: Mar 1, 2005
3. Mar 2, 2005

### Benny

Heh, I thought that homework exercises(mine is from an exercise booklet) wouldn't involve that much work.

I know that the following is a rather stupid question but I'd just like to make sure. For (b) as an example could I write something like let a = 2, b = -1, c = 4 and plug the values into back into (x,y,z) to get one solution? Anyway thanks for the help. Out of curiosity are you from Melbourne, Australia? :D

Last edited: Mar 2, 2005
4. Mar 2, 2005

### EvLer

Yeah, that would be one of many solutions.
Nope, quite far from it actually , right now I am going to school in US.
Our exercises come from Penney's textbook, who is a professor in our math department.

5. Mar 2, 2005

### Benny

Oh ok. Some of the questions from my booklet might have been taken from that textbook.

6. Mar 3, 2005

### mathwonk

interesting that a criterion for correctness was assumed to be lack of work involved.