# Iinearly independent solutions of homogeneous system of linear equations

Let's take an example of system of equations(actually they are all same)

x+2y+4z=0

2x+4y+8z=0

-3x-6y-12z=0

It is homogeneous system of linear equations. Therefore it has zero solution [0,0,0].

Rank of those coefficient matrix is found to be 1.

I read in textbook that it will have (n-r) linearly independent solutions where n= number of variables & r= rank of coefficient matrix.

So for above example it'll have (3-1)=2 linearly independent solutions.

I'll list some solutions out of infinitely many solutions:

[0,0,0]

[-6,1,1]

[-4,0,1]

[-2, 1,0] & so on.

Questions:

1)where are the 2 linearly independent solutions?

2)out of many solutions, 2 are found to be linearly independent. Is the rest of solutions Linearly dependent solutions?

3) what is actually meant by linearly dependence / independence?

If you reduce the coefficient matrix, you should find that the last 2 equations are a multiple of the first, so the system of equations reduces to $$x + 2y + 4z=0$$

As you noticed, if $$y=z=0$$, then $$x=0$$. So $$(0,0,0)$$ is a particular solution of the system.

Now we ask the question, what if $$y$$ and $$z$$ were not $$0$$? We can answer that question by making $$y\neq0$$, then $$z\neq0$$. You will notice that the vectors we end up with are NOT multiples of each other, so they are "linearly independent"

1) If $$y=c\neq0\in\mathbb{R}$$ and $$z=0$$, then $$x=-2c$$ and $$(-2c,c,0)=c(-2,1,0)$$ is a solution for any $$c\in\mathbb{R}$$.

2)If $$z=d\neq0\in\mathbb{R}$$ and $$y=0$$, then $$x=-4d$$ and $$(-4d,0,d)=d(-4,0,1)$$ is a solution for any $$d\in\mathbb{R}$$.

As mentioned earlier, there does not exist a $$c$$ and $$d$$ such that $$c(-2,1,0)+d(-4,0,1)=0$$. Therefore the vectors are not multiples of each other. This condition is the definition of linear independence

• Then I have infinite linearly independent solutions. But as I said in textbook it has given that it has only 2 linearly independent solutions – Ajay vishwanath Oct 5 '18 at 13:33
• The independent solutions are (-2,1,0) and (-4,0,1). The power of the fact they are linearly independent means their linear combinations form a solution space – NazimJ Oct 5 '18 at 13:39