Matrix one-parameter solution help

1. Dec 24, 2006

sara_87

hello people
i was wondering if anyone could help me with this question:

For what values of a and b does the system

ax +bz = 2
ax +ay +4z = 4
ay + 2z = b

have: (i) no solution
(ii) a unique solution
(iii) a one-parameter solution
(iv) a two-parameter solution?

this is what i did, i wrote it in a matrix form and i did gussian elimination, and across the diagonal i got a, a, and 1
i multiplied the digits on he main diagonal and i got a^2, i made that equal to 0... then i made this post
(basically i'm just doing random things with the numbers and i'm so stuck i don't know what to do! )

any tips would be really appreciated

(tomorrow is christmas )

2. Dec 24, 2006

arildno

Subtract the first from the second equation, and get:
ay+(4-b)z=2 (2*)

Subtract (2*) from your third equation, and get:
(b-2)z=(b-2)

Thus, you have the following system of equations to work with:
ax+bz=2
ay+(4-b)z=2
(b-2)z=(b-2)

3. Dec 24, 2006

sara_87

Thus, you have the following system of equations to work with:
ax+bz=2
ay+(4-b)z=2
(b-2)z=(b-2) (3*)

then i divided 3* by (b-2)

and i got
ax+bz=2
ay+(4-b)z=2
z=1

to work with...should i not do that last step?

and (i think you know this but) to get a no solution or a unique solution etc. we should use the determinant of the matrix which is the multiplication of the numbers on the main diagonal...i bet i'm talking rubbish...then we make the multiplication of the numbers on the main daigonal equal to zero, then i get stuck for real!

i just don't know what to do! lol

4. Dec 24, 2006

matt grime

You get 0,1 or infinitely many solutions if there are, well, 0, 1 or infinitely many solutions. So have a look at the solution set and let's see if when we can have 0,1 or infinitely many solutions.

We know (b-2)z=b-2. Now, you divided by b-2. That's fine *when it's allowed*. When is it allowed?

5. Dec 24, 2006

sara_87

When is it allowed?
'clueless'

6. Dec 24, 2006

cristo

Staff Emeritus
Suppose we have the equation ax=4. Then this gives the solution x=4/a iff a is not equal to zero. Does this help?

7. Dec 24, 2006

arildno

sara:
It is perfectly true that if the determinant of a quadratic matrix is non-zero, then we'll have a unique solution of the associated system of equations.

Thus, the "no solution" or "infinite solution" cases will have the value of the determinant equal to zero.
But do you see the problem here?

The problem is that if we just set the determinant equal to zero, gaining a relation between a and b, we haven't really DISTINGUISHED between the "no solution"/"infinite number of solutions" cases!!