1. ## Matrices With Variables

Ok, I'm supposed to determine the values of a for which the system has no solution, exactly one solution, or infinitely many solutions...

The system is:

x+y+7z=-7
2x+3y+17z=-16
x+2y+(a^2+1)z= 3a

So, I would have a matrix of

1 1 7 -7
2 3 17 -16
1 2 (a^2+1) 3a

I'm not sure how to solve for the system from this point using elementary row operations when it comes down to the a values...

2. Originally Posted by Hellreaver
Ok, I'm supposed to determine the values of a for which the system has no solution, exactly one solution, or infinitely many solutions...

The system is:

x+y+7z=-7
2x+3y+17z=-16
x+2y+(a^2+1)z= 3a

So, I would have a matrix of

1 1 7 -7
2 3 17 -16
1 2 (a^2+1) 3a

I'm not sure how to solve for the system from this point using elementary row operations when it comes down to the a values...
Do row operations to get echelon form.

Infinite solutions for values of a that give the form:

* * * | *
* * * | *
0 0 0 | 0

No solutions for values of a that give the form:

* * * | *
* * * | *
0 0 0 | Non-zero

Otherwise there's a unique solution.

This is a bit vague but you should understand what I mean if you understand this topic.

3. Ok, so I got:
1 1 7 -7
0 1 3 -2
0 0 1 (3a+9)/(a^2-9)

as my row echelon.

So, the solutions would be:
if
a=3: no solutions
a=-3: infinite solutions
a=/= +/-3: unique solution.

Is this correct?

4. Originally Posted by Hellreaver
Ok, so I got:
1 1 7 -7
0 1 3 -2
0 0 1 (3a+9)/(a^2-9)

as my row echelon.

So, the solutions would be:
if
a=3: no solutions
a=-3: infinite solutions
a=/= +/-3: unique solution.

Is this correct?
I'm not going to sift through the details but if your row echelon matrix is correct then you're correct in saying that there's no unique solution if a = +3 or -3. I assume you substituted each value of a into the original equations in order to test which of the cases infinite solutions and no solutions each value corresponded to.