# Thread: For what values of "a" is there one, none, or infinitely many solutions?

1. ## For what values of "a" is there one, none, or infinitely many solutions?

I normally wouldn't have too much trouble with something like this, but my skills regarding it are somewhat rusty.

For what values of $a$ does the system of equations
$x+ay+2z=0$
$-2x-ay+z=4$
$2ax+3a^2y+9z=4$
has one, none, or infinitely many solutions? Find all the possible solutions.

Right now, it's mainly that $a^2$ 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).

2. You can put this into a matrix:

$\left(
\begin{array}{ccc|c}
1 & a & 2 & 0 \\
-2 & -a & 1 & 4 \\
2 a & 3 a^2 & 9 & 4
\end{array}
\right)$

Try row-reducing this and see what kinds of assumptions you need for $a$ to get each case.

3. 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, $2ax+ 3a^2y+ 9z= 4: (3a^2- 2a)y+ (9- 4a)z= 4$.

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.

4. Originally Posted by HallsofIvy
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, $2ax+ 3a^2y+ 9z= 4: (3a^2- 2a)y+ (9- 4a)z= 4$.

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.
That looks alright, but you made a couple of mistakes: instead of $(2-a)y+5z=4$, it should be $ay+5z=4$, since the first equation is $x+ay+2=0$.

Also, instead of $(3a^2- 2a)y+ (9- 4a)z= 4$, it's supposed to be $a^2y+(9-4a)z=4$.

You forgot that the first equation had an $a$ in it.