Values of a for which the system has infinite number of solutions

[otapia13]

Find all values a (alpha) for which the resulting system has a) no solution, b) a unique solution c) infinitely many solutions.

x - y - az = -1
2x +y + az = -1
-x + y + (a^2 - 2)z = a + 2

attempt:

put it in matrix form and row-reduce

1 -1 -a : -1
2 1 a : -1
-1 1 a^2-2 : a+2
after row reducing I get:1 0 0 : 1
0 1 -2 : 1
0 0 a-2 : 1

so for a = 0 the system has a unique solution (you get x=1; y=0; and z=-1/2)
for a = 2 the system has no solution ( you get 0 0 0 : 1 on the bottom row)

so I have a values for a unique sol. and no sol. but I still need infinitely many solutions.

help.

EDIT: if I'm not mistaken really you get a unique solution for any real number, a, that is not 2 (not only 0 as stated above)
so how are you supposed to get infinitely many solutions?

 

[micromass]

Mod note: moved to homework forums

 

[SammyS]

Find all values a (alpha) for which the resulting system has a) no solution, b) a unique solution c) infinitely many solutions.

x - y - az = -1
2x +y + az = -1
-x + y + (a^2 - 2)z = a + 2

attempt:

put it in matrix form and row-reduce

1 -1 -a : -1
2 1 a : -1
-1 1 a^2-2 : a+2
after row reducing I get:


1 0 0 : 1
0 1 -2 : 1
0 0 a-2 : 1

so for a = 0 the system has a unique solution (you get x=1; y=0; and z=-1/2)
for a = 2 the system has no solution ( you get 0 0 0 : 1 on the bottom row)

so I have a values for a unique sol. and no sol. but I still need infinitely many solutions.

help.

EDIT: if I'm not mistaken really you get a unique solution for any real number, a, that is not 2 (not only 0 as stated above)
so how are you supposed to get infinitely many solutions?

If you add the first two equations together, you get 3x = -2 . So x = -2/3 no matter what the value of a is.

If a = 0, the solution that I get is: x = -2/3, y = 1/3. z = -1/2 .

Notice that in the third equation, z has a coefficient of (a² - 2) , not (a - 2) .