Matrices: find the values of a, b so that the system has....

[Vampire99]

Hey guys, this might seem like an easy problem, but I'm having a little difficulty tackling it...does anybody have any ideas?

Given real numbers a and b consider the system of linear equations:
5y+3z=1

ax+y-z=a-3

ax+2y+2z=a+2b-2

2ax-2y=2a+b

Find the values of a and b for which the equations have more than one solution, and find the general solution in this case.


Thx for your time.

 

[Guest]

You might have to revise your work with Matrices...

Remember that any linear system can be expressed in terms of matrices.

We then multiply this matrix by its inverse to get the identity matrix. The right handside will also be multiplied by the inverse. Then you get the answer.
Can you show us what you have done so far?

 

[stapel]

A good first step might be to rearrange the equations to get all the variables on the left-hand side and the constant terms on the right-hand side. Then write down the coefficient matrix.

Then consider the coefficient matrix for a dependent system. What sort of value does its determinant have?

There may be other methods for answering this exercise. If you are supposed to use one in particular, you will need to specify and/or show some work of your own.

Thank you.

Eliz.

 

[Vampire99]

Alrighty, this is what I've got so far...

original matrix:
0 5 3 | 1
a 1 -1 | a-3
a 2 2 | a+2b-2
2a -2 0 | 2a+b

through various ERO I end up with the matrix:
a 2 2 | a+2b-2
0 -1 -3 | -1-2b
0 0 14| 9b+10
0 0 0 | -16b/7 + 32/7

therefore,
-16b/7 +32/7 = 0 for the system to be consistant
i.e. b=2

However, if b=2, I get a unique solution for the system where 'a' can be any number except 0. And x,y,z have uniques values too as a result...

So I'm not sure how we can find the values of a and b for which the equations have more than one solution and the general solution as required, as I cannot seem to locate where any free variables come into it.

Thx again for your time guys.