Math Is Fun Forum

darfmore

Member

Registered: 2012-03-02

Posts: 4

Consistency/Inconsistancy of linear systems

Hey again all,

I'd just like a few ideas on how to go about solving this problem that I have been given.
The problem reads:

Consider the system of linear equations where a,b,c,d,e,f,g are real numbers and a =/= 0

ax + by +cz = 0
       dy + ez = 3
        fy + gz = 4

It then asks to write it in augmented matrix form (which is easy), from which we need to derive a condition that guarantees that the system has:
a) A unique solution
b) Infinitely many solutions

Now I have no idea as to how to go about answering this.
I have figured that there are 3 possibilities for the solution/s:
- Infinite
- Unique
- Inconsistent (No solution)

For it to be inconsistent then I think that d and e must be proportional to f and g. Because then if you subtract the two rows from each other you're left with an equation that reads 0 = 1, therefore inconsistent. I think that this is the only way that the third option can be arrived at.

I'm not sure about the other two, though. Any suggestions about how to go about this?
Any help would be greatly appreciated.

Thanks!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Consistency/Inconsistancy of linear systems

Hi;

In matrix notation we have:

In order for that system to not have a solution the matrix

can not be invertible. That would mean the determinant of it would be 0. Do you agree?

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Consistency/Inconsistancy of linear systems

hi darfmore

ax + by +cz = 0 .........................(1)
       dy + ez = 3.........................(2)
        fy + gz = 4.........................(3)

If (2) and (3) have a solution, substitute the values for y and z into (1) and that fixes x.

So when will (2) and (3) have a solution?

Draw the straight line graphs for (2) and (3).

If (2) is parallel to (3) then there's no solution.  That is what you have said with:

For it to be inconsistent then I think that d and e must be proportional to f and g. Because then if you subtract the two rows from each other you're left with an equation that reads 0 = 1, therefore inconsistent.

But it is possible that (2) and (3) represent the same line, in which case there are infinitely many solutions; just pick a value of y and z from the line.  So your 'proportional' argument needs a small modification to allow for the 'same' line case.

If (2) isn't parallel to (3) then the lines cross once and you have a unique solution for y and z, leading to a single value for x.

So the condition you need to consider is the value of dg - ef.

If it's not zero you have a unique solution.

If it is zero you need to check if

If yes, there are an infinity of solutions.

If no, then there are no solutions.

note:  This is equivalent to trying to find the inverse matrix.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob