# Homework Help: Linear algebra - set of solutions

1. Nov 11, 2012

### peripatein

Hello,

Is it true that the following system of linear equations would always have a single solution (i.e. would never have an infinite number of solutions nor none) for any value of λ?

λx + 3y -z = 1
x + 2y -z = 2
-λx + y + 2z = -1

May someone kindly confirm?

2. Nov 11, 2012

### micromass

How did you conclude that? What did you try?

3. Nov 11, 2012

### peripatein

Via a series of elementary operations I got:
lambda*x +7y = 1
(1/2)x + y -(1/2)z = 1
4y + z = 0

So, rank of the matrix is equal to the number of columns, 3, is it not? Wouldn't that always have a solution, for any lambda that is?

4. Nov 11, 2012

### peripatein

Alas, I am still experiencing difficulties trying to find the values of lambda for which number of solutions of the system of equations is: (i) 1; (ii) infinite; (iii) 0
Having substituted the equations into a matrix and performed a series of elementary operations on the rows, I have found out that there is a single solution for lambda different than 3 and 7/6. However, when lambda is 3 the system still has a single solution, hence a contradiction!
What am I doing wrong? I have tried various row operations several times, and each time upon substitution of those values for lambda (for which it couldn't possibly have a single solution) the system still has a single solution!

5. Nov 11, 2012

### Ray Vickson

For some values of λ the determinant of the coefficient matrix vanishes. What does that tell you?

RGV

6. Nov 11, 2012

### peripatein

We haven't dealt with determinants so I wouldn't know, and even if I did I am not permitted to use employ any knowledge thereof. Could you please just tell me what it is I might be doing wrong? For hours I have tried reducing the matrix and each time the lambda's I got for which the number of solutions would be one, ended up, upon substitution, yielding a solution! So I am pretty stuck and would appreciate some more practical guidance, please.