Solving for infinitely many solutions in a linear system with matrices

[Bardagath]

Consider the system

6x-3y=9
ax+y=-3

where a is a real number.

For which value of a has this system infinitely many solutions?

My intuition tells me this should be a straightforward problem but I am not sure how to get started with it. My effort so far has been to put the system in a matrix form to solve for a but I don't see how this enables us to find a value of a which gives infinitely many solutions.

Could someone tell me what it means for this system to have infinitely many solutions? Is the number 0 involved in here somewhere? Do we need to put x+y into an indentity form, so that any real number will work?

- Bardagath

 

[Anonymous217]

In order for this system to have infinitely many solutions, the two equations would have to produce the exact same curve (well technically, it doesn't have to, but it's the easiest way to solve the problem). So try to make both equations able to equal each other.
Multiply the second equation by -3 on both sides, getting -3ax -3y = 9.
Set them equal, 6x - 3y = -3ax - 3y.
-3ax = 6x
a = -2

You can tell that the problem wanted this method because everything conveniently cancels out. I'm guessing this is an SAT question.

 

[Bardagath]

Anonymous217,

Your method is most inspiring and it helped me conquer this problem. What do you think about this method:

put both equations 6x-3y=9 and ax+y=-3 into an augmented matrix form

------> 6 | -3 : 9 (-a) multiply row 1 by -a
A | 1 : 6 (6) and add to 6 times row 2
-------------
0 | (3a+6) : -9a-18

If you solve 3a+6 = -9a-18 you get an answer of -2 but I would like to know if this is wrong, should it not be (3a+6)y = -9a-18? Due to the term being in the Y column of the matrix

EDIT: I tried solving the system (3a+6)y = -9a-18 and the solution set was x=3, y=-3 a=0. Can someone help me on how to attack this problem in matrix form?