Finding Single Solution for ax+y+z=0 System

[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 that's 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

 

[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.

 

[Mark44]

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

Your equations are equivalent to

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?

 

[Dell]

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

 

[Mark44]

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.

 

[Dell]

dont know what to do with this, i get a=a-1 which can't be, can you see any possible solutions for this problem

 

[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