# Thread: Gauss' Method Proof

1. ## Gauss' Method Proof

This is a really basic question. I want to prove the following statment:

If a linear system is changed to another by multiplying both sides of an equation by a nonzero constant, then the two systems have the same solution.

I know this is very obvious and seemingly pointless to prove. However, I never have been able to tackle the concept of proof and I can rarely figure out how to begin on things like this.

This is my linear system:

$a_{1,1}x_1+a_{1,2}x_2+...+a_{1,n}x_n=d_1$

$a_{2,1}x_1+a_{2,2}x_2+...+a_{2,n}x_n=d_2$

$a_{m,1}x_1+a_{m,2}x_2+...+a_{m,n}x_n=d_m$

The solution set is an n-tuple ${s_1,s_2....s_n}$. Where would I go from here?

2. I wrote out an attempt to prove thie statement in my original post.

$a_{1,1}x_1+a_{1,2}x_2+...+a_{1,n}x_n=d_1$

$a_{2,1}x_1+a_{2,2}x_2+...+a_{2,n}x_n=d_2$

$a_{m,1}x_1+a_{m,2}x_2+...+a_{m,n}x_n=d_m$

The solution set is an n-tuple ${s_1,s_2....s_n}$.

Since the n-tuple is a solution to the linear system, I can choose the linear combination $a_{2,1}x_1+a_{2,2}x_2+...+a_{2,n}x_n=d_2$ and write the true statement $a_{2,1}S_1+a_{2,2}S_2+...+a_{2,n}S_n=d_2$. Multiplying both sides by a nonzero constant $C$ gives $Ca_{2,1}S_1+Ca_{2,2}S_2+...+Ca_{2,n}S_n=Cd_2$. But this is just the requirement for the linear combination $Ca_{2,1}x_1+Ca_{2,2}x_2+...+Ca_{2,n}x_n=Cd_2$ to have a solution set $S_1,...,S_n$. Therefore, the solution set is unchanged.