# Linear algebra

1. Jan 15, 2009

### Dell

given this system, to solve, find the possible values for a so there is a single solution and that x=y

ax+y+z=0
x+(a+1)y+z=0
x+y+z=a

i put the equations into a matrix and got to

a-1 0 0 | -a
0 a 0 | -a
0 0 1 | a+1+(a/(a-1))

so
(a-1)x=-a
ay=-a
z=a+1+(a/(a-1))

does this mean that x,y and a are all equal to zero, because thats the only answer i can see here that answers to all the conditions,

(0-1)0=-0
0*0=-0
z=1

doesnt seem right to me, can someone please check it for me

2. Jan 15, 2009

### MathematicalPhysicist

x=y, then:
(a+1)x+z=0
(a+2)x+z=0
2x+z=a
(a+1)x=(a+2)x
this occurs iff x=0=y
and z=a.

3. Jan 15, 2009

### Staff: Mentor

x = -a/(a - 1) (so a cannot be 1)
y = -a/a (so a cannot be 0)
z = a + 1 + a/(a - 1) (a cannot be 1)

You should check that these values actually are solutions of your original system of equations. Substitute these values in your original system. If they are solutions, you should get three true statements.

I suspect that they aren't, because the problem stated that you were to show there was only one solution. With the solution you showed, there are an infinite number of solutions, one for each valid value of a.

Are you sure you have posted the exact problem description?

4. Jan 15, 2009

### Dell

i think what you are missing is that x=y, that should limit the answers somehow, but i posted the exact question

5. Jan 15, 2009

### Staff: Mentor

You should still check that your solution for x, y, and z actually satisfies your original system of equations. If so, then you can set x equal to y to find a value for a.

6. Jan 16, 2009

### Dell

dont know what to do with this, i get a=a-1 which cant be, can you see any possible solutions for this problem

7. Jan 16, 2009

### Дьявол

$$x=\frac{a}{1-a}$$
y=-z
$$z=\frac{a^2-a+2}{a-1}$$
a≠1
The only possible solutions are:
x=y=z=a=0