# Homework Help: Linear Algebra Concept Question

1. Jul 27, 2008

### LHC

The problem statement, all variables and given/known data

List the conditions of "a" under which the system:

x + ay - z = 1
-x + (a-2) y + z = -1
2x + 2y (a-2) z = 1

i) Has no solutions:
ii) Has a unique solution:
iii) Has infinite solutions.

The attempt at a solution

Well, I changed the matrix into reduced row-echelon form and I found:

x = (a-1)/a
y = 0
z = -1/a

The textbook gives the following answers:
i) No solutions: a = 0 (which I understand...because otherwise there would be an undefined x and z)

ii) Unique solution: a not= 0 and a not= 1, then unique solution of:
x = (a-1)/a
y = 0
z = -1/a (same as above)

iii) Infinite solutions: if a = 1, then solutions of x = -t, y = t, z = -1

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I don't understand the third case for infinite solutions. If a = 1, then wouldn't: x = 0, y = 0, z = -1?

Also, what is the general strategy for solving such problems? (Many of these types of problems are given in the section of the textbook that I'm going through, yet no example was ever given.) I mean, I understand how to get a unique solution, and how to get no solution, but I'm not sure how you would deduce the case for infinite solutions.

2. Jul 27, 2008

### konthelion

A Linear System may have
1. A unique solution - the system is consistent
2. Infinite solutions - the system is consistent
3. No solutions - the system is inconsistent

To distinguish between a unique solution and infinite solutions:
If there are no basic variables, then the system has a unique solution.
If there are >= 1 free variables, the solutions has infinite solutions.

In
x = -t, y = t, z = -1
which are the basic, and free variables? Since there are at least 1 free variable, then the system has infinite solutions since t can be any number.

3. Jul 28, 2008

### konthelion

Sorry, this is a typo, it should be no free variables.

4. Jul 28, 2008

### HallsofIvy

You mean 2x+ 2y+ (a-2)z= 1, don't you?

Be careful here. One of the equations reduces to 2(a-1)y= 0 which gives y= 0 provided a is not equal to 1. If a= 1, that equation is 0= 0 which is true no matter what y is. That's the case where there are an infinite number of solutions.

5. Jul 28, 2008

### LHC

Thanks, konthelion and HallsofIvy, for your help. HallsofIvy, when you said "Be careful here. One of the equations reduces to 2(a-1)y= 0 which gives y= 0 provided a is not equal to 1. If a= 1, that equation is 0= 0 which is true no matter what y is. That's the case where there are an infinite number of solutions.", I looked back on the steps that I used to reduce the matrix and you're right. Thank you for clarifying this for me; now I understand the rest of these problems.