You can put this into a matrix:
Try row-reducing this and see what kinds of assumptions you need for to get each case.
I normally wouldn't have too much trouble with something like this, but my skills regarding it are somewhat rusty.
For what values of does the system of equations
has one, none, or infinitely many solutions? Find all the possible solutions.
Right now, it's mainly that in the third equation that has me a tad confused on how to handle it. This shouldn't be very hard, but, as I said, my memory on this sort of thing is a bit rusty (and muddled by other topics).
Or, more "unsophisticatedly" (my favorite kind- although it usually involves more work) solve the equations directly.
For example, if we add twice the first equation, 2x+ 2y+ 4z= 0, to the second equation, -2x- ay+ z= 4, we eliminate x: (2-a)y+ 5z= 4.
We can also eliminate x by adding -2a times the first equation, -2ax- 2ay- 4az= 0, to the third equation, .
Now that you have two equations in y and z only, you could combine them to eliminate, say, z, leaving an equation for y only (and a, of course). Solving that equation for y will involve dividing one expression in a by another. There will be no solution if and only if the denominator is 0 but the numerator is not.