What do, unique, infinite and no solutions mean?

[you0006]

Can you relate this to matrices, or tell me what this means to the determinant. I've noticed like nearly each VCAA paper has at least 1 question that looks like that, i just have no idea how to solve them

[tony3272]

I like to look at these questions as the actual lines themselves.

A unique solution means that they cross each other once. This will only occur if the gradient of the two lines are different, as parallel lines will never cross.

The next situation is no solutions. This means that the two lines are parallel to each other and so they will never cross or intersect each other.

Lastly, infinitely many solutions. This is the one that i think confuses people the most, but what it essentially means is that the lines intersect each other infinitely many times. Hence this can only occur if the two lines are actually the same line. (i.e if you draw a line on top of another it technically touches itself an infinite amount of times)

[b^3]

EDIT: With the determinate, when you setup the equation, if the determinate = 0 then there is no sols, i.e. parallel so infinite sols or no sols. When it does not equal 0 then there are sols.
e.g. you have 2x+3y=1 and x+5y=2


Original:
This may help too, it's something I pm'ed someone one day.

There is two ways to do these questions. The first is the matrix method where you setup a mtrix equation, take the determinate (ad-bc) of the sqaure matrix and let it equal 0 then solve for m.

I prefer the second method though.
What you need to know is that (this is for two linear equations)
1) For a unique solution - there will be only one intersection point. This means that the gradients of the two lines cannot be the same
2) Infinite solutions - there will be a solution on every point on both lines i.e. the two lines are the same line. the gradients of the two lines are the same and share all points in common
3) No solutions - as it says no solutions - the gradients are the same but they do not have a point in common i.e. they are parallel and are not the same line.

So the way to go about it is to rearrange the equations into y=mx+c form. From there you can take the gradients of both and let them equal each other. Depending on what you are looking for determines where you go next.
1) A unique Solution - solved values of m will give when the gradients are the same, we are looking for when they are not the same. So these values are what they cannot be. e.g. if m=a,b then the values we are looking for is R\{a,b}
2) Infinite solutions - gradients will be the same so that is what the solved values of m will be. Now to test wether they are the same lines, subsitute the values of m into the equation and check wether they give the same equation.
3) No solutions - solved values of m is for the same gradient. Then plug into oriignal equations to check for when the two equations are not the same

Example
Find the values of m for when the equations −3x + (m+1) y = 5 and 8y − (m+ 3) x = (2m+ 4) have
i) a unqiue solution
ii) infinte solutions
iii) no solutions
Firstly rearrange to y=mx+c
(m+1)y=5+3x

8y=(m+3)x+(2m+4)

now solve for when m is the same





So that is when the lines are parallel (i.e. same gradients)
Equating the gradients
So check wether -7 or 3 give two different equations
m=-7,
equation 1
equation 2
So the two equations are different
m=3,
equation 1
equation 2
So they are the same equation
i) a unqiue solution
different gradients, m is a member of R\{-7,3}
ii) infinte solutions
same gradients, same line
m=3
iii) no solutions
same gradient, different lines
m=-7

I hope that helps

[brightsky]

unique solution means det =/= 0
for both no solutions and infinite solutions, det = 0

to clarify let us imagine two random lines:
ax + by = e ----> gradient = -a/b
cx + dy = f ----> gradient = -c/d

we know that if -a/b = -c/d, then the equations either have no solutions or infinite solutions, since they are either parallel to each other, or are exactly the same line. rearranging the equation...-ad = -bc, ad - bc = 0. [HINT HINT SIMILARITIES WITH DETERMINANT]

we also know that if -a/b =/= -c/d, then the equations have only one solution, since lines can only ever INTERSECT at one point. rearranging equation...-ad =/= -bc, ad - bc =/= 0. that is ad - bc is a non-zero number.

now look at the matrix version:

[a,b;c,d][x;y] = [e;f]
to find [x;y] we must take the multiply both sides by [a,b;c,d]^(-1) = 1/determinant [d,-b;-c,a]
but for the inverse matrix to exist, it is clear that the determinant must be a non-zero number. that is ad - bc =/= 0. [RELATE TO ABOVE]
if the determinant is a non-zero number, we end up solving the matrix, and there will only be one solution.
however if the determinant is 0, that is ad - bc = 0, then there is no inverse, hence we cannot multiply both sides by the inverse matrix and hence there will effectively be no way in using matrix methods, in which case we can only rely on intuitive methods to find out whether the answer is no or infinite solutions.