Proving Solutions to Homogeneous Linear Systems

[Jacobpm64]

Homework Statement

Consider the homogeneous system of linear equations
$$ax + by = 0$$
$$cx + dy = 0$$
Prove that if $$ad - bc \not= 0$$, then $$x = 0,y=0$$ is the only solution to the system.

The Attempt at a Solution

First, I tried rewriting the system of equations to get $$y = -\frac{a}{b} x$$ and $$y = -\frac{c}{d} x$$. This would have probably helped me in the proof, but I realized that I may not be able to divide by $$b$$ and $$d$$ because they may be $$0$$.

Maybe I could use the contrapositive to prove this. Proving the statement "If $$ad - bc = 0$$, then $$x = 0, y = 0$$ is not the only solution to the system. I'd have to show that there are more solutions. I am not sure how to do this though. Although, this seems like the easiest way to prove it.



Can anyone give me some help? Thanks in advance.

 

[Mathdope]

Are you familiar with determinants yet?

 

[Jacobpm64]

No, this is the beginning of the Linear course. We didn't do much more than an introduction to matrices and Reduced Echelon Form for Gauss-Jordan Elimination.

Oh I'm sorry, I posted the question incorrectly. It was an if and only if proof, and I did that direction pretty easily.

The implication I actually need help on is the opposite direction.

If x = 0, y = 0 is the only solution to the system, then $$ad - bc \not= 0$$.

 

[Mathdope]

As you said if ad = bc, then you should be able to show that (0,0) is not the only solution. Look at what happens in the original system if you assume this, and you then multiply the top equation by d and the bottom equation by b.

 

[HallsofIvy]

Homework Statement

Consider the homogeneous system of linear equations
$$ax + by = 0$$
$$cx + dy = 0$$
Prove that if $$ad - bc \not= 0$$, then $$x = 0,y=0$$ is the only solution to the system.

The Attempt at a Solution

First, I tried rewriting the system of equations to get $$y = -\frac{a}{b} x$$ and $$y = -\frac{c}{d} x$$. This would have probably helped me in the proof, but I realized that I may not be able to divide by $$b$$ and $$d$$ because they may be $$0$$.

Then say "if b is not 0" and "if d is not 0" and see what you get. Then go back and say "if b= 0" or "if d= 0" and see what you get.

Maybe I could use the contrapositive to prove this. Proving the statement "If $$ad - bc = 0$$, then $$x = 0, y = 0$$ is not the only solution to the system. I'd have to show that there are more solutions. I am not sure how to do this though. Although, this seems like the easiest way to prove it.

Can anyone give me some help? Thanks in advance.

No, that's not the contrapositive. You want to prove "If ad-bc is not 0 then x= 0, y= 0 is the only solution". The contrapositive of that is "if x= 0, y= 0 is not the only solution then ad- bc= 0." I think you would find that harder. Have you considered just going ahead and solving the system? Try starting with ax+ by= 0 and multiplying both sides by d: adx+ bdy= 0. Now take cx+ dy= 0 and multiply both sides by -b: -bcx- bdy= 0. Adding those two equations will eliminate y. Do you see why you need "ad- bc not equal to 0"? What would happen if ad- bc= 0?

 

[Jacobpm64]

Mathdope:

assume ad = bc.

multiplying the top equation by d and the bottom by b:
$$d(ax + by) = 0$$
$$b(cx + dy) = 0$$

$$adx + bdy = 0$$
$$bcx + bdy = 0$$

Now I'm sure there's something I can do with ad and bc being the coefficients of x.. since they're equal.

I'm just not sure though.

 

[Mystic998]

If ad = bc, then adx + bdy = 0 and bcx + bdy = 0 are the same equation. What does that say about the solutions to the system?

 

[Jacobpm64]

The lines coincide so there are infinitely-many solutions.

So that's all I need to do for that proof?

 

[Mathdope]

Mathdope:

assume ad = bc.

multiplying the top equation by d and the bottom by b:
$$d(ax + by) = 0$$
$$b(cx + dy) = 0$$

$$adx + bdy = 0$$
$$bcx + bdy = 0$$

Now I'm sure there's something I can do with ad and bc being the coefficients of x.. since they're equal.

I'm just not sure though.

$$adx + bdy = 0$$
$$bcx + bdy = 0$$

Notice that these two imply that

$$adx - bcx = 0$$

Edit: notice that you don't need to assume that ad = bc to perform the multiplications. Just do them and then see what happens IF ad = bc.

 

[Mystic998]

Oh, sorry, I didn't read carefully. You either need to show that if there's a solution other than (0,0) then ad = bc or follow what Mathdope suggested to the logical conclusion.

 

[Jacobpm64]

Mathdope:

Okay.. so if i factor out an x:

(ad - bc)x = 0

since ad = bc,
we get 0*x = 0.

But that doesn't tell us anything, does it?

I'm still confused :(

 

[Mathdope]

If ad = bc, then x can be anything. If ad - bc is nonzero, what must x be? Reread the original problem.

 

[Jacobpm64]

the original problem is... "If x = 0, y = 0 is the only solution, then $$ad - bc \not= 0$$."


To answer your question:

If ad - bc is nonzero then x must be zero.

I didn't think I could assume that ad - bc is nonzero in this proof though.

Since I'm assuming ad - bc = 0... Or similarly, i can assume that x = 0, y = 0 is the only solution.

 

[Mathdope]

If ad - bc = 0 then x can have any value. This means that (x,y) = (0,0) is not the only solution. If you think of x as a parameter t, then any ordered pair of the form (t,-(a/b)t) is a solution (for nonzero b). If b = 0, then (0,t) is a solution for all t. (Notice that if ad = bc, then a/b = c/d when b and d aren't zero).