# Thread: Basic Linear Algebra Proof

1. ## Basic Linear Algebra Proof

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.

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.

2. Originally Posted by Jacobpm64
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.
If the determinant, $ad-bc$, is non-zero then this equation has exactly one solution i.e. the trivial solution.

But, it seems you never learned how determinants correspond to linear equations, thus we will do it another way.

You can write,
$adx+bdy=0$
$bcx+bdy=0$
Subtract two equations,
$(ad - bc) x = 0$
Since $ad-bc\not =0$ it means $x = 0$ which in turn implies $y=0$. Thus, there is only the trivial solution.

3. oh, i screwed up.. it was an if and only if..

I proved that part pretty easily..

the part i actually needed help with is the other direction.

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